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

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

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

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

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

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

- 1 + 2 cos (2 x) = 0

Move

1

to the right side

2 cos (2 x) = 1

Divide both sides by

2

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

General solutions for

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

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

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

Solve

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

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

tan (x) = 12

2 + 3 cos^{2} (θ) = sin^{2} (θ) + 4 cos (θ)

arctan (4 x) + arctan (6 x) = \frac{π}{2}

x, y = \frac{1}{tan ( x \cdot π )}

cos (x) = sin (\frac{7 π}{6})

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