What is the general solution for -2sin(x)+1=0 ?

The general solution for -2sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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-2sin(x)+1=0

- 2 sin (x) + 1 = 0

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

- 2 sin (x) + 1 = 0

Move

1

to the right side

- 2 sin (x) = - 1

Divide both sides by

- 2

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (2 θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

2 - 2 cos^{2} (\frac{x}{2}) = 2 sin^{2} (x)

3 tanh^{2} (x) = 5 sech (x) + 1

tan (θ) = \frac{5}{7}

(cot (x) - 1) (3 tan (x) + 1) = 0

What is the general solution for -2sin(x)+1=0 ?
The general solution for -2sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin