What is the general solution for 1+4*sin(x)=2 ?

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

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1+4*sin(x)=2

1 + 4 \cdot sin (x) = 2

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

Degrees

x = 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 165.52248 \dots^{\circ} + 36 0^{\circ} n

Solution steps

1 + 4 sin (x) = 2

Move

1

to the right side

1 + 4 sin (x) = 2

Subtract

1

from both sides

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

Simplify

4 sin (x) = 1

4 sin (x) = 1

Divide both sides by

4

4 sin (x) = 1

Divide both sides by

4

\frac{4 sin ( x )}{4} = \frac{1}{4}

Simplify

sin (x) = \frac{1}{4}

sin (x) = \frac{1}{4}

Apply trig inverse properties

sin (x) = \frac{1}{4}

General solutions for

sin (x) = \frac{1}{4}

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

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

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

Show solutions in decimal form

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

cos (θ) = \frac{7}{8}

sec (x) (2 sin (x) + 1) - 2 (2 sin (x) + 1) = 0

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

1 = sin (x) + 3 cos (x)

1 + sin (x) = 2 cos^{2} (x), 0 \leq x < 5

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