Find the general solutions of each of the following trigonometric equations:

a) 3sinēx+ cosx- 1 = 0

b) cos 6θ+ cos 2θ= 0

Thank you for your help :)

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- Jun 5th 2006, 05:30 PMLilDragonflyPlease help!
Find the general solutions of each of the following trigonometric equations:

a) 3sinē*x*+ cos*x*- 1 = 0

b) cos 6*θ*+ cos 2*θ*= 0

Thank you for your help :) - Jun 5th 2006, 05:53 PMmalaygoelQuote:

Originally Posted by**LilDragonfly**

using qudratic formula you will get cosx =1, -2/3

hence, x = [2n*pi ]and [2n*pi + cos inverse of (-2/3)] - Jun 5th 2006, 06:31 PMThePerfectHackerQuote:

Originally Posted by**LilDragonfly**

$\displaystyle \cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) $

Thus,

$\displaystyle \cos 6x+\cos 2x=0$

Use the formula,

$\displaystyle 2\cos 4x\cos 2x=0$

Thus,

$\displaystyle \left\{ \begin{array}{c} \cos 4x=0\\\cos 2x=0$ - Jun 9th 2006, 07:21 PMLilDragonflyQuote:

Originally Posted by**malaygoel**

- Jun 10th 2006, 03:22 AMSoroban
Hello, LilDragonfly!

Quote:

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

This is what Malaygoel suggested . . .

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

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

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

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

. . and has two equations to solve:

. . . . . $\displaystyle \cos x - 1 \:=\:0\quad\Rightarrow\quad\cos x = 1\quad\Rightarrow\quad x = 0$

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

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