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 \leq 2 π

x = \frac{3 π}{2}

Solution steps

sin (x) + 1 = 0, 0 \leq x \leq 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 \leq 2 π

x = \frac{3 π}{2}

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

12 tan (x) + 5 = 5 tan (x) + 5

4 cos (x) + 3 sin (x) = 0

2 sin (4 x) - 1 = 0, 0 < x \leq 2

5 cos (θ) = - 5 sin (- θ)

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