Solve for x
x=\frac{-2\left(\sin(\theta )\right)^{5}+1}{3}
Solve for θ
\theta =-\arcsin(\frac{2^{\frac{4}{5}}\sqrt[5]{1-3x}}{2})+2\pi n_{1}+\pi \text{, }n_{1}\in \mathrm{Z}
\theta =\arcsin(\frac{2^{\frac{4}{5}}\sqrt[5]{1-3x}}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }x\geq -\frac{1}{3}\text{ and }x\leq 1
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3x=1-2\left(\sin(\theta )\right)^{5}
Subtract 2\left(\sin(\theta )\right)^{5} from both sides.
3x=-2\left(\sin(\theta )\right)^{5}+1
The equation is in standard form.
\frac{3x}{3}=\frac{-2\left(\sin(\theta )\right)^{5}+1}{3}
Divide both sides by 3.
x=\frac{-2\left(\sin(\theta )\right)^{5}+1}{3}
Dividing by 3 undoes the multiplication by 3.
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