What is the general solution for 5sin(θ)+3=0 ?

The general solution for 5sin(θ)+3=0 is θ=-0.64350…+2pin,θ=pi+0.64350…+2pin

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5sin(θ)+3=0

5 sin (θ) + 3 = 0

θ = - 0.64350 \dots + 2 πn, θ = π + 0.64350 \dots + 2 πn

Degrees

θ = - 36.86989 \dots^{\circ} + 36 0^{\circ} n, θ = 216.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (θ) + 3 = 0

Move

3

to the right side

5 sin (θ) + 3 = 0

Subtract

3

from both sides

5 sin (θ) + 3 - 3 = 0 - 3

Simplify

5 sin (θ) = - 3

5 sin (θ) = - 3

Divide both sides by

5

5 sin (θ) = - 3

Divide both sides by

5

\frac{5 sin ( θ )}{5} = \frac{- 3}{5}

Simplify

sin (θ) = - \frac{3}{5}

sin (θ) = - \frac{3}{5}

Apply trig inverse properties

sin (θ) = - \frac{3}{5}

General solutions for

sin (θ) = - \frac{3}{5}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{3}{5}) + 2 πn, θ = π + arcsin (\frac{3}{5}) + 2 πn

θ = arcsin (- \frac{3}{5}) + 2 πn, θ = π + arcsin (\frac{3}{5}) + 2 πn

Show solutions in decimal form

θ = - 0.64350 \dots + 2 πn, θ = π + 0.64350 \dots + 2 πn

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

cos (x) + cos (2 x) = - 1

cos (θ) = \frac{6}{3 5}

3 cos (x) = 2 cos^{2} (x)

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

What is the general solution for 5sin(θ)+3=0 ?
The general solution for 5sin(θ)+3=0 is θ=-0.64350…+2pin,θ=pi+0.64350…+2pin