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

The general solution for 2sin(6x)+2sin(2x)=0 is x=pin,x=(pi+2pin)/2 ,x=(3pi+4pin)/4 ,x=(pi+4pin)/4

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

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

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

Solution steps

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

Let:

u = 2 x

2 sin (3 u) + 2 sin (u) = 0

Rewrite using trig identities

8 sin (u) - 8 sin^{3} (u) = 0

Solve by substitution

sin (u) = 0, sin (u) = - 1, sin (u) = 1

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

sin (u) = 1 : u = \frac{π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = 2 x

2 x = 2 πn : x = πn

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

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

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

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

tan^{2} (2 x) - 1 = sec (2 x)

- 1.92 = cos (x)

\frac{sin ^{2} ( x )}{4} = \frac{cos ^{2} ( x )}{2}

θ, 2 sin (θ) = tan (θ)

5 cot^{2} (x) + 8 cot (x) + 3 = 0

What is the general solution for 2sin(6x)+2sin(2x)=0 ?
The general solution for 2sin(6x)+2sin(2x)=0 is x=pin,x=(pi+2pin)/2 ,x=(3pi+4pin)/4 ,x=(pi+4pin)/4