## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 280: 44

#### Answer

The solutions for this problem is $x=$ {$90^\circ+n\pi, n\in Z$} and {$45^\circ+n\pi, n\in Z$}

#### Work Step by Step

$$\sin2x=2\cos^2 x$$ We would use the identity $\sin2x=2\sin x\cos x$, which means $$2\sin x\cos x=2\cos^2 x$$ $$\sin x\cos x=\cos^2 x$$ $$\sin x\cos x-\cos^2 x=0$$ $$\cos x(\sin x-\cos x)=0$$ $$\cos x=0$$ or $$\sin x=\cos x$$ * For $\cos x=0$, $x=90^\circ$ To cover the whole set of solutions, we see that for every $180^\circ$ from the position $x=90^\circ$ forward or backward, $\cos x$ again equals $0$. Therefore, the full set of solutions in this case is $x=$ {$90^\circ+n\pi, n\in Z$} * For $\sin x=\cos x$ In this case, we consider that $\cos x\ne0$, since we have considered the case above. Therefore, we can divide both sides by $\cos x$ $$\frac{\sin x}{\cos x}=\frac{\cos x}{\cos x}$$ $$\tan x=1$$ $$x=45^\circ$$ Again, to cover the whole set of solutions, we need to add or subtract $180^\circ$ with the basic angle $x=45^\circ$, since for every $180^\circ$ forward or backward, $\tan x$ again equals $1$. The full set of solutions is $x=$ {$45^\circ+n\pi, n\in Z$} In conclusion, the solutions for this problem is $x=$ {$90^\circ+n\pi, n\in Z$} and {$45^\circ+n\pi, n\in Z$}

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