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

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

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

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

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

Solution steps

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

Rewrite using trig identities

2 sin^{2} (x) + 2 cos (x) sin (x) = 0

Factor

2 sin^{2} (x) + 2 cos (x) sin (x) : 2 sin (x) (sin (x) + cos (x))

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

Combine all the solutions

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

5 sin (x) cos (x) + 4 cos (x) = 0

2 sin (x) = 4 cos (x)

6 arcsin (x) = 2 π

8 cos (x) = 0

2 sin (x) \cdot tan (x) + 5 sin (x) = - 2 cos (x)

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