Solving Trigonometric Equations

**anna_sims** · October 31st 2006, 06:15 PM

Solving trigonometric equations

I need help on solving these problems within specific intervals. I am having trouble solving for x the most. Any help is again appreciated.

1)

5-7sin x = 2cos^2 x x = [ -360, 450 degrees ]

2)

2csc^2 x = 3cot^2 x – 1 x = [ 0, 2pi ]

This is just a sampling of my homework, but it will help me a ton if someone can go through these correctly.

**Soroban** · October 31st 2006, 08:26 PM

Hello, Anna!

$1)\;5 - 7\sin x \:= \:2\cos^2x\qquad [-360^o,\,450^o]$

Replace $\cos^2x$ with $1 - \sin^2x$

. . $5 - 7\sin x\:=\:2(1 - \sin^2x)\quad\Rightarrow\quad 5 - 7\sin x \:=\:2 - 2\sin^2x$

We have a quadratic: . $2\sin^2x - 7\sin x + 3 \:=\:0$

. . which factors: . $(\sin x - 3)(2\sin x - 1)\:=\:0$

Then we have: . $\sin x - 3\:=\:0\quad\Rightarrow\quad \sin x \:=\:3$ ... no real root

And: . $2\sin x - 1\:=\:0\quad\Rightarrow\quad \sin x \,= \,\frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:-330^o,\,-210^o,\,30^o,\,150^o,\,390^o}$

$2)\;2\csc^2x \:= \:3\cot^2x - 1\qquad [0,\,2\pi ]$

Replace $\cos^2x$ with $\csc^2x - 1$

. . $2\csc^2x \:=\:3(\csc^2x - 1) - 1\quad\Rightarrow\quad 2\csc^2x \:=\:3\csc^2x - 3 - 1$

We have: . $\csc^2x\:=\:4\quad\Rightarrow\quad \csc x\:=\:\pm2\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\,\frac{5\pi}{6},\,\frac{7\pi}{ 6},\,\frac{11\pi}{6}}$

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