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

The general solution for 4sin(x)+1=-2 is No Solution for x\in\mathbb{R}

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4sin(x)+1=-2

4 sin (x^{\circ}) + 1 = - 2

No Solution for x \in R

Solution steps

4 sin (x^{\circ}) + 1 = - 2

Move

1

to the right side

4 sin (x^{\circ}) + 1 = - 2

Subtract

1

from both sides

4 sin (x^{\circ}) + 1 - 1 = - 2 - 1

Simplify

4 sin (x^{\circ}) = - 3

4 sin (x^{\circ}) = - 3

Divide both sides by

4

4 sin (x^{\circ}) = - 3

Divide both sides by

4

\frac{4 sin ( x ^{\circ} )}{4} = \frac{- 3}{4}

Simplify

sin (x^{\circ}) = - \frac{3}{4}

sin (x^{\circ}) = - \frac{3}{4}

Apply trig inverse properties

sin (x^{\circ}) = - \frac{3}{4}

General solutions for

sin (x^{\circ}) = - \frac{3}{4}

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

x^{\circ} = arcsin (- \frac{3}{4}) + 2 πn, x^{\circ} = π + arcsin (\frac{3}{4}) + 2 πn

x^{\circ} = arcsin (- \frac{3}{4}) + 2 πn, x^{\circ} = π + arcsin (\frac{3}{4}) + 2 πn

No Solution for x \in R

(2 cos (x) + 1) (3 tan (x) - 1) = 0

α, cos^{2} (α) + cos^{2} (β) = 1

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

sin (x + 3 0^{\circ}) = \frac{1}{2}

sin (θ) = 21

What is the general solution for 4sin(x)+1=-2 ?
The general solution for 4sin(x)+1=-2 is No Solution for x\in\mathbb{R}