# solving trig equations

• Nov 14th 2007, 02:50 PM
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solving trig equations
Solve each equation. 0 < x < 2pi
1. 3sinx = cosx
2. sin2x = sinx

Find the points of intersection of each pair of curves in the given interval.
y=sin2x, y=sinx
0 < x < 2pi
• Nov 14th 2007, 03:29 PM
Soroban
Hello, checkmarks!

Quote:

Solve each equation. . $0 \leq x \leq 2\pi$

$1)\;\;3\sin x \:= \:\cos x$

Divide both sides by $\cos x\!:\;\;\frac{3\sin x}{\cos x} \:=\:\frac{\cos x}{\cos x}\quad\Rightarrow\quad 3\tan x \:=\:1$

Then: . $\tan x\:=\:\frac{1}{3}\quad\Rightarrow\quad x \:=\:\tan^{-1}\left(\frac{1}{3}\right)\quad\Rightarrow\quad\bo xed{ x\;\approx\;0.32175,\;3.46334}$

Quote:

$2)\;\;\sin2x \:= \:\sin x$

We have: . $2\sin x\cos x \:=\:\sin x$

. . $2\sin x\cos x - \sin x \:=\:0\quad\Rightarrow\quad \sin x(2\cos x - 1)\:=\:0$

Then:. . $\sin x \:=\: 0 \quad\Rightarrow\quad\boxed{ x \:=\:0,\:\pi,\:2\pi}$
. . . . . . $2\cos x - 1 \:= \:0 \quad\Rightarrow\quad \cos x \:=\: \frac{1}{2} \quad\Rightarrow\quad \boxed{x\:=\: \frac{\pi}{3},\;\frac{5\pi}{3}}$

Quote:

Find the points of intersection of this pair of curves in the given interval.

. . $\begin{array}{ccc}y &=&\sin2x \\y &= &\sin x\end{array}\qquad 0\,<\,x\,<\,2\pi$

This has the same equation as the preceding problem.

Answers: . $\boxed{x \:=\:\pi,\;\frac{\pi}{3},\;\frac{5\pi}{3}}$