What is the general solution for 1/(cos^2(x))= 2/(3-3sin(x)) ?

The general solution for 1/(cos^2(x))= 2/(3-3sin(x)) is x= pi/6+2pin,x=(5pi)/6+2pin

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Popular Trigonometry >

1/(cos^2(x))= 2/(3-3sin(x))

Solution

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

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

Subtract

\frac{2}{3 - 3 sin ( x )}

from both sides

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

Simplify

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

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

Factor

3 - 3 sin (x) : - 3 (sin (x) - 1)

3 - 3 sin (x)

Factor out common term

- 3

= - 3 (sin (x) - 1)

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

Least Common Multiplier of

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

cos^{2} (x), - 3 (sin (x) - 1)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos^{2} (x)

- 3 (sin (x) - 1)

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

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

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

For

\frac{1}{cos ^{2} ( x )} :

multiply the denominator and numerator by

- 3 (sin (x) - 1)

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

For

\frac{2}{- 3 ( sin ( x ) - 1 )} :

multiply the denominator and numerator by

cos^{2} (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

Apply the fraction rule:

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

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

Expand

- 3 (- 1 + sin (x)) : 3 - 3 sin (x)

- 3 (- 1 + sin (x))

Apply the distributive law:

a (b + c) = ab + a c

a = - 3, b = - 1, c = sin (x)

= - 3 (- 1) + (- 3) sin (x)

Apply minus-plus rules

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

= 3 \cdot 1 - 3 sin (x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin (x)

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

3 - 2 = 1

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

= - 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 = \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

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 1/(cos^2(x))= 2/(3-3sin(x)) ?
The general solution for 1/(cos^2(x))= 2/(3-3sin(x)) is x= pi/6+2pin,x=(5pi)/6+2pin