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

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

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

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

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

Solution steps

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

Subtract

3

from both sides

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

Rewrite using trig identities

- 2 - 4 cos (2 x) = 0

Move

2

to the right side

- 4 cos (2 x) = 2

Divide both sides by

- 4

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

General solutions for

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

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

Solve

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

Solve

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

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

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

sin^{2} (a) + \frac{1}{sec ( a )} = \frac{5}{4}

\frac{csc ( x )}{cot ( x )} = 2

sin (4 x) + 6 cos (2 x) = 0

sec (θ) + tan (θ) = 0

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