What is the general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 ?

The general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 is x= pi/6+(pin)/4

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sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) = \frac{1}{2}

x = \frac{π}{6} + \frac{πn}{4}

Solution steps

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) = \frac{1}{2}

Subtract

\frac{1}{2}

from both sides

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) - \frac{1}{2} = 0

Simplify

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) - \frac{1}{2} : \frac{2 3 sin ^{2} ( 2 x ) - sin ( 4 x ) - 3}{2 3}

\frac{2 3 sin ^{2} ( 2 x ) - sin ( 4 x ) - 3}{2 3} = 0

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

23 sin^{2} (2 x) - sin (4 x) - 3 = 0

Rewrite using trig identities

- tan (4 x) - 3 = 0

Move

3

to the right side

- tan (4 x) = 3

Divide both sides by

- 1

tan (4 x) = - 3

General solutions for

tan (4 x) = - 3

4 x = \frac{2 π}{3} + πn

Solve

4 x = \frac{2 π}{3} + πn : x = \frac{π}{6} + \frac{πn}{4}

x = \frac{π}{6} + \frac{πn}{4}

x, sin (x + x y) = 0

sin (x) = - 3 cos (x)

- 2 sin^{2} (θ) + cos (θ) + 8 = 6

sin^{3} (t) = \frac{1}{8}

2 cos (t) \cdot sin (t) = cos (t)

What is the general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 ?
The general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 is x= pi/6+(pin)/4