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

The general solution for 2sin(x)+1=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6

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2sin(x)+1=0,0<= x<= 2pi

2 sin (x) + 1 = 0, 0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

2 sin (x) + 1 = 0, 0 \leq x \leq 2 π

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

9 csc^{2} (θ) - 25 = 0

16 cos^{2} (θ) - 25 = 0

(2 cos (x) + 3) (2 sin (x) - 1) = 0

(tan (x) + 3) (2 sin (x) - 1) = 0

2 cos (3 θ) + 1 = 0

What is the general solution for 2sin(x)+1=0,0<= x<= 2pi ?
The general solution for 2sin(x)+1=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6