What is the general solution for 2cos(x)+3tan(x)=3sec(x) ?

The general solution for 2cos(x)+3tan(x)=3sec(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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Popular Trigonometry >

2cos(x)+3tan(x)=3sec(x)

Solution

2 cos (x) + 3 tan (x) = 3 sec (x)

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) + 3 tan (x) = 3 sec (x)

Subtract

3 sec (x)

from both sides

2 cos (x) + 3 tan (x) - 3 sec (x) = 0

Express with sin, cos

2 cos (x) + 3 \cdot \frac{sin ( x )}{cos ( x )} - 3 \cdot \frac{1}{cos ( x )} = 0

Simplify

2 cos (x) + 3 \cdot \frac{sin ( x )}{cos ( x )} - 3 \cdot \frac{1}{cos ( x )} : \frac{2 cos ^{2} ( x ) + 3 sin ( x ) - 3}{cos ( x )}

2 cos (x) + 3 \cdot \frac{sin ( x )}{cos ( x )} - 3 \cdot \frac{1}{cos ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )} = \frac{3 sin ( x )}{cos ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( x ) \cdot 3}{cos ( x )}

3 \cdot \frac{1}{cos ( x )} = \frac{3}{cos ( x )}

3 \cdot \frac{1}{cos ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{cos ( x )}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{cos ( x )}

= 2 cos (x) + \frac{3 sin ( x )}{cos ( x )} - \frac{3}{cos ( x )}

Combine the fractions

\frac{3 sin ( x )}{cos ( x )} - \frac{3}{cos ( x )} : \frac{3 sin ( x ) - 3}{cos ( x )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{3 sin ( x ) - 3}{cos ( x )}

= 2 cos (x) + \frac{3 sin ( x ) - 3}{cos ( x )}

Convert element to fraction:

2 cos (x) = \frac{2 cos ( x ) cos ( x )}{cos ( x )}

= \frac{2 cos ( x ) cos ( x )}{cos ( x )} + \frac{sin ( x ) \cdot 3 - 3}{cos ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 cos ( x ) cos ( x ) + sin ( x ) \cdot 3 - 3}{cos ( x )}

2 cos (x) cos (x) + sin (x) \cdot 3 - 3 = 2 cos^{2} (x) + 3 sin (x) - 3

2 cos (x) cos (x) + sin (x) \cdot 3 - 3

2 cos (x) cos (x) = 2 cos^{2} (x)

2 cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= 2 cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (x)

= 2 cos^{2} (x) + 3 sin (x) - 3

= \frac{2 cos ^{2} ( x ) + 3 sin ( x ) - 3}{cos ( x )}

\frac{2 cos ^{2} ( x ) + 3 sin ( x ) - 3}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cos^{2} (x) + 3 sin (x) - 3 = 0

Subtract

3 sin (x)

from both sides

2 cos^{2} (x) - 3 = - 3 sin (x)

Square both sides

(2 cos^{2} (x) - 3)^{2} = (- 3 sin (x))^{2}

Subtract

(- 3 sin (x))^{2}

from both sides

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x) = 0

Factor

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x) : (2 cos^{2} (x) - 3 + 3 sin (x)) (2 cos^{2} (x) - 3 - 3 sin (x))

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x)

Rewrite

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x)

(2 cos^{2} (x) - 3)^{2} - (3 sin (x))^{2}

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x)

Rewrite

9

3^{2}

= (2 cos^{2} (x) - 3)^{2} - 3^{2} sin^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

3^{2} sin^{2} (x) = (3 sin (x))^{2}

= (2 cos^{2} (x) - 3)^{2} - (3 sin (x))^{2}

= (2 cos^{2} (x) - 3)^{2} - (3 sin (x))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos^{2} (x) - 3)^{2} - (3 sin (x))^{2} = ((2 cos^{2} (x) - 3) + 3 sin (x)) ((2 cos^{2} (x) - 3) - 3 sin (x))

= ((2 cos^{2} (x) - 3) + 3 sin (x)) ((2 cos^{2} (x) - 3) - 3 sin (x))

Refine

= (2 cos^{2} (x) + 3 sin (x) - 3) (2 cos^{2} (x) - 3 sin (x) - 3)

(2 cos^{2} (x) - 3 + 3 sin (x)) (2 cos^{2} (x) - 3 - 3 sin (x)) = 0

Solving each part separately

2 cos^{2} (x) - 3 + 3 sin (x) = 0 or 2 cos^{2} (x) - 3 - 3 sin (x) = 0

2 cos^{2} (x) - 3 + 3 sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

2 cos^{2} (x) - 3 + 3 sin (x) = 0

Rewrite using trig identities

- 3 + 2 cos^{2} (x) + 3 sin (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 3 + 2 (1 - sin^{2} (x)) + 3 sin (x)

Simplify

- 3 + 2 (1 - sin^{2} (x)) + 3 sin (x) : 3 sin (x) - 2 sin^{2} (x) - 1

- 3 + 2 (1 - sin^{2} (x)) + 3 sin (x)

Expand

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (x)

= - 3 + 2 - 2 sin^{2} (x) + 3 sin (x)

Add/Subtract the numbers:

- 3 + 2 = - 1

= 3 sin (x) - 2 sin^{2} (x) - 1

= 3 sin (x) - 2 sin^{2} (x) - 1

- 1 - 2 sin^{2} (x) + 3 sin (x) = 0

Solve by substitution

- 1 - 2 sin^{2} (x) + 3 sin (x) = 0

Let:

sin (x) = u

- 1 - 2 u^{2} + 3 u = 0

- 1 - 2 u^{2} + 3 u = 0 : u = \frac{1}{2}, u = 1

- 1 - 2 u^{2} + 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + 3 u - 1 = 0

Solve with the quadratic formula

- 2 u^{2} + 3 u - 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 3, c = - 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

3^{2} - 4 (- 2) (- 1) = 1

3^{2} - 4 (- 2) (- 1)

Apply rule

- (- a) = a

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 ( - 2 )}, u_{2} = \frac{- 3 - 1}{2 ( - 2 )}

u = \frac{- 3 + 1}{2 ( - 2 )} : \frac{1}{2}

\frac{- 3 + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 1}{- 2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

= \frac{- 2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 1}{2 ( - 2 )} : 1

\frac{- 3 - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 1}{- 2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 1

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = 1

sin (x) = \frac{1}{2}, sin (x) = 1

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

2 cos^{2} (x) - 3 - 3 sin (x) = 0 : x = \frac{3 π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

2 cos^{2} (x) - 3 - 3 sin (x) = 0

Rewrite using trig identities

- 3 + 2 cos^{2} (x) - 3 sin (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 3 + 2 (1 - sin^{2} (x)) - 3 sin (x)

Simplify

- 3 + 2 (1 - sin^{2} (x)) - 3 sin (x) : - 2 sin^{2} (x) - 3 sin (x) - 1

- 3 + 2 (1 - sin^{2} (x)) - 3 sin (x)

Expand

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (x)

= - 3 + 2 - 2 sin^{2} (x) - 3 sin (x)

Add/Subtract the numbers:

- 3 + 2 = - 1

= - 2 sin^{2} (x) - 3 sin (x) - 1

= - 2 sin^{2} (x) - 3 sin (x) - 1

- 1 - 2 sin^{2} (x) - 3 sin (x) = 0

Solve by substitution

- 1 - 2 sin^{2} (x) - 3 sin (x) = 0

Let:

sin (x) = u

- 1 - 2 u^{2} - 3 u = 0

- 1 - 2 u^{2} - 3 u = 0 : u = - 1, u = - \frac{1}{2}

- 1 - 2 u^{2} - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} - 3 u - 1 = 0

Solve with the quadratic formula

- 2 u^{2} - 3 u - 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 3, c = - 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) (- 1) = 1

(- 3)^{2} - 4 (- 2) (- 1)

Apply rule

- (- a) = a

= (- 3)^{2} - 4 \cdot 2 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 1}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 1}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 1}{2 ( - 2 )} : - 1

\frac{- ( - 3 ) + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 + 1}{- 2 \cdot 2}

Add the numbers:

3 + 1 = 4

= \frac{4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 3 ) - 1}{2 ( - 2 )} : - \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 - 1}{- 2 \cdot 2}

Subtract the numbers:

3 - 1 = 2

= \frac{2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = - \frac{1}{2}

sin (x) = - 1, sin (x) = - \frac{1}{2}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 cos (x) + 3 tan (x) = 3 sec (x)

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{6} + 2 πn :

True

\frac{π}{6} + 2 πn

Plug in

n = 1

\frac{π}{6} + 2 π 1

For

2 cos (x) + 3 tan (x) = 3 sec (x)

plug in

x = \frac{π}{6} + 2 π 1

2 cos (\frac{π}{6} + 2 π 1) + 3 tan (\frac{π}{6} + 2 π 1) = 3 sec (\frac{π}{6} + 2 π 1)

Refine

3.46410 \dots = 3.46410 \dots

\Rightarrow True

Check the solution

\frac{5 π}{6} + 2 πn :

True

\frac{5 π}{6} + 2 πn

Plug in

n = 1

\frac{5 π}{6} + 2 π 1

For

2 cos (x) + 3 tan (x) = 3 sec (x)

plug in

x = \frac{5 π}{6} + 2 π 1

2 cos (\frac{5 π}{6} + 2 π 1) + 3 tan (\frac{5 π}{6} + 2 π 1) = 3 sec (\frac{5 π}{6} + 2 π 1)

Refine

- 3.46410 \dots = - 3.46410 \dots

\Rightarrow True

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

2 cos (x) + 3 tan (x) = 3 sec (x)

plug in

x = \frac{π}{2} + 2 π 1

2 cos (\frac{π}{2} + 2 π 1) + 3 tan (\frac{π}{2} + 2 π 1) = 3 sec (\frac{π}{2} + 2 π 1)

Refine

\infty = \infty

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

2 cos (x) + 3 tan (x) = 3 sec (x)

plug in

x = \frac{3 π}{2} + 2 π 1

2 cos (\frac{3 π}{2} + 2 π 1) + 3 tan (\frac{3 π}{2} + 2 π 1) = 3 sec (\frac{3 π}{2} + 2 π 1)

Refine

\infty = - \infty

\Rightarrow False

Check the solution

\frac{7 π}{6} + 2 πn :

False

\frac{7 π}{6} + 2 πn

Plug in

n = 1

\frac{7 π}{6} + 2 π 1

For

2 cos (x) + 3 tan (x) = 3 sec (x)

plug in

x = \frac{7 π}{6} + 2 π 1

2 cos (\frac{7 π}{6} + 2 π 1) + 3 tan (\frac{7 π}{6} + 2 π 1) = 3 sec (\frac{7 π}{6} + 2 π 1)

Refine

0 = - 3.46410 \dots

\Rightarrow False

Check the solution

\frac{11 π}{6} + 2 πn :

False

\frac{11 π}{6} + 2 πn

Plug in

n = 1

\frac{11 π}{6} + 2 π 1

For

2 cos (x) + 3 tan (x) = 3 sec (x)

plug in

x = \frac{11 π}{6} + 2 π 1

2 cos (\frac{11 π}{6} + 2 π 1) + 3 tan (\frac{11 π}{6} + 2 π 1) = 3 sec (\frac{11 π}{6} + 2 π 1)

Refine

0 = 3.46410 \dots

\Rightarrow False

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)+3tan(x)=3sec(x) ?
The general solution for 2cos(x)+3tan(x)=3sec(x) is x= pi/6+2pin,x=(5pi)/6+2pin