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

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

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-sin(2x)-sin(x)=0

- sin (2 x) - sin (x) = 0

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

Solution steps

- sin (2 x) - sin (x) = 0

Rewrite using trig identities

- sin (x) - 2 cos (x) sin (x) = 0

Factor

- sin (x) - 2 cos (x) sin (x) : - sin (x) (2 cos (x) + 1)

- sin (x) (2 cos (x) + 1) = 0

Solving each part separately

sin (x) = 0 or 2 cos (x) + 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

2 cos (x) + 1 = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

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

sin^{2} (t) + 2 cos (t) = 1

6 tan (θ) - 2 cot (θ) = 0

- 2 sin^{2} (x) = 7 cos (x) + 2

2 sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

cos (x) (2 cos (x) + 1) = 3

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