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

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

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-sin(2x)(2)=0

- sin (2 x) (2) = 0

x = πn, x = \frac{π}{2} + πn

Solution steps

- sin (2 x) (2) = 0

Divide both sides by

- 2

sin (2 x) = 0

General solutions for

sin (2 x) = 0

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

cos (5 x - 21) = sin (13 - 3 x)

sin (θ) = \frac{5}{3}

cos (2 θ) + 14 sin^{2} (θ) = 10, 0 < θ < 2 π

14 sin^{2} (x) + 5 sin (x) - 1 = 0

sin (x) = \frac{3}{5} sin (x + \frac{π}{2}) + cos (π - x)

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