# solving trig equations

• Nov 14th 2007, 01:50 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, 02:29 PM
Soroban
Hello, checkmarks!

Quote:

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

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

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

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

Quote:

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

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

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

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

Quote:

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

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

This has the same equation as the preceding problem.

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