I am practice about find the x as solution of sin⁡(2x3+36∘)=12 Between 0∘ and 360∘ Here is my work so far, sin⁡(2x3+36∘)=sin⁡30∘ I try using arcsin 2x3+36∘=30∘ 2x3=−6∘ x=−9∘ I am stucked here.

We can use that sin⁡A=sin⁡B⇔A=B+2πk∨A=π−B+2πk with k∈Z.

Solved: "I am practice about find the x as..."

Swebaceacichegh

2022-01-24

I am practice about find the x as solution of

\sin (\frac{2 x}{3} + 36^{\circ}) = \frac{12}{}

Between

0^{\circ} and 360^{\circ}

Here is my work so far,

\sin (\frac{2 x}{3} + 36^{\circ}) = {\sin 30}^{\circ}

I try using arcsin

\frac{2 x}{3} + 36^{\circ} = 30^{\circ}

\frac{2 x}{3} = - 6^{\circ}

x = - 9^{\circ}

I am stucked here.

Micah May

Beginner2022-01-25Added 11 answers

We can use that

\sin A = \sin B \Leftrightarrow A = B + 2 π k \lor A = π - B + 2 π k

with

k \in Z

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