Solve for Q
Q=\frac{\sqrt{2}}{2\cos(\theta )}
\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}
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2Q\cos(\theta )=\sqrt{2}
Multiply both sides of the equation by 2.
2\cos(\theta )Q=\sqrt{2}
The equation is in standard form.
\frac{2\cos(\theta )Q}{2\cos(\theta )}=\frac{\sqrt{2}}{2\cos(\theta )}
Divide both sides by 2\cos(\theta ).
Q=\frac{\sqrt{2}}{2\cos(\theta )}
Dividing by 2\cos(\theta ) undoes the multiplication by 2\cos(\theta ).
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