What is the general solution for sin^2(6x)+sin^2(3x)=0 ?

The general solution for sin^2(6x)+sin^2(3x)=0 is x=(2pin)/3 ,x=(pi+2pin)/3

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sin^2(6x)+sin^2(3x)=0

sin^{2} (6 x) + sin^{2} (3 x) = 0

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

Solution steps

sin^{2} (6 x) + sin^{2} (3 x) = 0

Let:

u = 3 x

sin^{2} (2 u) + sin^{2} (u) = 0

Rewrite using trig identities

sin^{2} (u) + 4 cos^{2} (u) sin^{2} (u) = 0

Factor

sin^{2} (u) + 4 cos^{2} (u) sin^{2} (u) : sin^{2} (u) (4 cos^{2} (u) + 1)

sin^{2} (u) (4 cos^{2} (u) + 1) = 0

Solving each part separately

sin^{2} (u) = 0 or 4 cos^{2} (u) + 1 = 0

sin^{2} (u) = 0 : u = 2 πn, u = π + 2 πn

4 cos^{2} (u) + 1 = 0 :

No Solution

Combine all the solutions

u = 2 πn, u = π + 2 πn

Substitute back

u = 3 x

3 x = 2 πn : x = \frac{2 πn}{3}

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

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

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

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

1 - tan^{2} (x) = a^{2} + b^{2}

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

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

What is the general solution for sin^2(6x)+sin^2(3x)=0 ?
The general solution for sin^2(6x)+sin^2(3x)=0 is x=(2pin)/3 ,x=(pi+2pin)/3