What is the general solution for sin(3x)+sin(x)=0,0<= x<= 2pi ?

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

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sin(3x)+sin(x)=0,0<= x<= 2pi

sin (3 x) + sin (x) = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π, x = 2 π

Solution steps

sin (3 x) + sin (x) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

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

Solving each part separately

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

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

sin (2 x) = 0, 0 \leq x \leq 2 π : x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = 2 π

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π, x = 2 π

4 sin (θ) + 3 = 1

25 cot^{2} (θ) - 1 = 0

cos (a) = \frac{5}{13}

sec (2 x) - 2 = 0

2 sin^{2} (θ) + sin (θ) = 1

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