What is the general solution for 2sin(θ)+6=-sqrt(3)+6 ?

The general solution for 2sin(θ)+6=-sqrt(3)+6 is θ=(4pi)/3+2pin,θ=(5pi)/3+2pin

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2sin(θ)+6=-sqrt(3)+6

2 sin (θ) + 6 = - 3 + 6

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

2 sin (θ) + 6 = - 3 + 6

Subtract

6

from both sides

2 sin (θ) + 6 - 6 = - 3 + 6 - 6

Simplify

2 sin (θ) = - 3

Divide both sides by

2

sin (θ) = - \frac{3}{2}

General solutions for

sin (θ) = - \frac{3}{2}

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin (\frac{x}{2} - \frac{π}{3}) = \frac{1}{2}, 0 \leq x \leq 2 π

4 cot (x) + 1 = 0

tan (x) + sin (x) tan (x) = 0

cosh (x) = 4

2 cos^{2} (θ) + 3 cos (θ) - 2 = 0

What is the general solution for 2sin(θ)+6=-sqrt(3)+6 ?
The general solution for 2sin(θ)+6=-sqrt(3)+6 is θ=(4pi)/3+2pin,θ=(5pi)/3+2pin