What is the general solution for 4-3sin(θ)=7 ?

The general solution for 4-3sin(θ)=7 is θ=(3pi)/2+2pin

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4-3sin(θ)=7

4 - 3 sin (θ) = 7

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

4 - 3 sin (θ) = 7

Move

4

to the right side

4 - 3 sin (θ) = 7

Subtract

4

from both sides

4 - 3 sin (θ) - 4 = 7 - 4

Simplify

- 3 sin (θ) = 3

- 3 sin (θ) = 3

Divide both sides by

- 3

- 3 sin (θ) = 3

Divide both sides by

- 3

\frac{- 3 sin ( θ )}{- 3} = \frac{3}{- 3}

Simplify

sin (θ) = - 1

sin (θ) = - 1

General solutions for

sin (θ) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (x) cos (x) = - \frac{1}{4}, 0 \leq x \leq 2 π

tan (x) = (\frac{4}{3})

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

x, sin (x) - cos (x) = 0

4 sin^{2} (x) + 4 cos (2 x) = 1

What is the general solution for 4-3sin(θ)=7 ?
The general solution for 4-3sin(θ)=7 is θ=(3pi)/2+2pin