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

The general solution for sin(2x)-sin(x)=0,(0,2pi) is x=pi,x= pi/3 ,x=(5pi)/3

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

sin (2 x) - sin (x) = 0, (0, 2 π)

x = π, x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

sin (2 x) - sin (x) = 0, (0, 2 π)

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, (0, 2 π) : x = π

2 cos (x) - 1 = 0, (0, 2 π) : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = π, x = \frac{π}{3}, x = \frac{5 π}{3}

sin^{4} (x) + cos^{4} (x) = \frac{7}{9}

tan (\frac{θ}{5}) + 3 = 0

3 tan (x) - 3 = 0, 0 \leq x < 2 π

sec^{2} (θ) = 2

sin^{2} (θ) = 8 - 2 cos^{2} (θ)

What is the general solution for sin(2x)-sin(x)=0,(0,2pi) ?
The general solution for sin(2x)-sin(x)=0,(0,2pi) is x=pi,x= pi/3 ,x=(5pi)/3