Please help!

**LilDragonfly** · June 5th 2006, 05:30 PM

Find the general solutions of each of the following trigonometric equations:
a) 3sin²x + cos x - 1 = 0
b) cos 6θ + cos 2θ = 0

Thank you for your help

**malaygoel** · June 5th 2006, 05:53 PM

Originally Posted by LilDragonfly

Find the general solutions of each of the following trigonometric equations:
a) 3sin²x + cos x - 1 = 0
b) cos 6θ + cos 2θ = 0

Thank you for your help

a) change (sinx)^2 into 1- (cosx)^2 you will get a quadratic in cosx
using qudratic formula you will get cosx =1, -2/3
hence, x = [2n*pi ]and [2n*pi + cos inverse of (-2/3)]

**ThePerfectHacker** · June 5th 2006, 06:31 PM

Originally Posted by LilDragonfly

b) cos 6θ + cos 2θ = 0

Use the formula,
$\cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$
Thus,
$\cos 6x+\cos 2x=0$
Use the formula,
$2\cos 4x\cos 2x=0$
Thus,
$\left\{ \begin{array}{c} \cos 4x=0\\\cos 2x=0$

**LilDragonfly** · June 9th 2006, 07:21 PM

Originally Posted by malaygoel

a) change (sinx)^2 into 1- (cosx)^2 you will get a quadratic in cosx
using qudratic formula you will get cosx =1, -2/3
hence, x = [2n*pi ]and [2n*pi + cos inverse of (-2/3)]

Hi again, would you be able to write this in equation format so that it is easier for me to understand? Thank you

**Soroban** · June 10th 2006, 03:22 AM

Hello, LilDragonfly!

$a)\;3\sin^2x + \cos x - 1 \:=\:0$

This is what Malaygoel suggested . . .

We have: . $3\underbrace{\sin^2x} + \cos x - 1 \;= \;0$

Then: . $3\overbrace{(1 - \cos^2x)} + \cos x - 1 \;= \;0$

And we have a quadratic: . $3\cos^2x - \cos x - 2 \;= \;0$

. . which factors: . $(\cos x - 1)(3\cos x + 2)\;=\;0$

. . and has two equations to solve:

. . . . . $\cos x - 1 \:=\:0\quad\Rightarrow\quad\cos x = 1\quad\Rightarrow\quad x = 0$
. . . . . $3\cos x + 2 \:=\:0\quad\Rightarrow\quad \cos x = -\frac{2}{3}\quad\Rightarrow\quad x = \cos^{-1}\left(-\frac{2}{3}\right)$

If they want all possible solutions, you can generalize . . . right?

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