What is the general solution for 3sin^2(2x)+7cos(2x)-3=0 ?

The general solution for 3sin^2(2x)+7cos(2x)-3=0 is x= pi/4+pin,x=(3pi)/4+pin

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Popular Trigonometry >

3sin^2(2x)+7cos(2x)-3=0

Solution

3 sin^{2} (2 x) + 7 cos (2 x) - 3 = 0

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

3 sin^{2} (2 x) + 7 cos (2 x) - 3 = 0

Rewrite using trig identities

- 3 + 3 sin^{2} (2 x) + 7 cos (2 x)

Use the Pythagorean identity:

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

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

= - 3 + 3 (1 - cos^{2} (2 x)) + 7 cos (2 x)

Simplify

- 3 + 3 (1 - cos^{2} (2 x)) + 7 cos (2 x) : 7 cos (2 x) - 3 cos^{2} (2 x)

- 3 + 3 (1 - cos^{2} (2 x)) + 7 cos (2 x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (2 x)

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

Multiply the numbers:

3 \cdot 1 = 3

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

= - 3 + 3 - 3 cos^{2} (2 x) + 7 cos (2 x)

- 3 + 3 = 0

= 7 cos (2 x) - 3 cos^{2} (2 x)

= 7 cos (2 x) - 3 cos^{2} (2 x)

- 3 cos^{2} (2 x) + 7 cos (2 x) = 0

Solve by substitution

- 3 cos^{2} (2 x) + 7 cos (2 x) = 0

Let:

cos (2 x) = u

- 3 u^{2} + 7 u = 0

- 3 u^{2} + 7 u = 0 : u = 0, u = \frac{7}{3}

- 3 u^{2} + 7 u = 0

Solve with the quadratic formula

- 3 u^{2} + 7 u = 0

Quadratic Equation Formula:

For

a = - 3, b = 7, c = 0

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

7^{2} - 4 (- 3) \cdot 0 = 7

7^{2} - 4 (- 3) \cdot 0

Apply rule

- (- a) = a

= 7^{2} + 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 7^{2} + 0

7^{2} + 0 = 7^{2}

= 7^{2}

Apply radical rule:

assuming

a \geq 0

= 7

u_{1, 2} = \frac{- 7 \pm 7}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 7 + 7}{2 ( - 3 )}, u_{2} = \frac{- 7 - 7}{2 ( - 3 )}

u = \frac{- 7 + 7}{2 ( - 3 )} : 0

\frac{- 7 + 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 + 7}{- 2 \cdot 3}

Add/Subtract the numbers:

- 7 + 7 = 0

= \frac{0}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{- 6}

Apply the fraction rule:

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

= - \frac{0}{6}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 7 - 7}{2 ( - 3 )} : \frac{7}{3}

\frac{- 7 - 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 - 7}{- 2 \cdot 3}

Subtract the numbers:

- 7 - 7 = - 14

= \frac{- 14}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 14}{- 6}

Apply the fraction rule:

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

= \frac{14}{6}

Cancel the common factor:

2

= \frac{7}{3}

The solutions to the quadratic equation are:

u = 0, u = \frac{7}{3}

Substitute back

u = cos (2 x)

cos (2 x) = 0, cos (2 x) = \frac{7}{3}

cos (2 x) = 0, cos (2 x) = \frac{7}{3}

cos (2 x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos (2 x) = 0

General solutions for

cos (2 x) = 0

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

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

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

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos (2 x) = \frac{7}{3} :

No Solution

cos (2 x) = \frac{7}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(2x)+7cos(2x)-3=0 ?
The general solution for 3sin^2(2x)+7cos(2x)-3=0 is x= pi/4+pin,x=(3pi)/4+pin