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

The general solution for sin(x)+1=0,0<= x<2pi is x=(3pi)/2

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

sin (x) + 1 = 0, 0 \leq x < 2 π

x = \frac{3 π}{2}

Solution steps

sin (x) + 1 = 0, 0 \leq x < 2 π

Move

1

to the right side

sin (x) = - 1

General solutions for

sin (x) = - 1

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

θ, sin^{2} (θ) = \frac{1}{4}

cot (θ) = 3, 0 \leq θ < 2 π

(2 cos (2 θ)) = 2

5 sin^{2} (x) = 9 tan (x)

cos (θ) = \frac{5}{4}

What is the general solution for sin(x)+1=0,0<= x<2pi ?
The general solution for sin(x)+1=0,0<= x<2pi is x=(3pi)/2