What is the general solution for 1+4sin(θ)=0 ?

The general solution for 1+4sin(θ)=0 is θ=-0.25268…+2pin,θ=pi+0.25268…+2pin

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1+4sin(θ)=0

1 + 4 sin (θ) = 0

θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

Degrees

θ = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 194.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

1 + 4 sin (θ) = 0

Move

1

to the right side

1 + 4 sin (θ) = 0

Subtract

1

from both sides

1 + 4 sin (θ) - 1 = 0 - 1

Simplify

4 sin (θ) = - 1

4 sin (θ) = - 1

Divide both sides by

4

4 sin (θ) = - 1

Divide both sides by

4

\frac{4 sin ( θ )}{4} = \frac{- 1}{4}

Simplify

sin (θ) = - \frac{1}{4}

sin (θ) = - \frac{1}{4}

Apply trig inverse properties

sin (θ) = - \frac{1}{4}

General solutions for

sin (θ) = - \frac{1}{4}

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

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

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

sin (x) - sin (2 x) = 0, x \leq 2 π, 0

- sin (x) - 2 cot (x) + 5 csc (x) = 0

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

cos (x) = \frac{π}{3}

What is the general solution for 1+4sin(θ)=0 ?
The general solution for 1+4sin(θ)=0 is θ=-0.25268…+2pin,θ=pi+0.25268…+2pin