Trigonometric equation

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• Sep 21st 2009, 02:18 AM
deltaxray
Trigonometric equation
Can someone please help me solve the following trig equations please? Thanks

1-tan^2(x)/1+tan^2(x) + cos(x) = 0

(sqrt(3)+1)cos^2(x) - 1 = (sqrt(3)-1)sin(x)cos(x)

Thanks heaps
• Sep 21st 2009, 03:39 AM
mathaddict
Quote:

Originally Posted by deltaxray
Can someone please help me solve the following trig equations please? Thanks

1-tan^2(x)/1+tan^2(x) + cos(x) = 0

(sqrt(3)+1)cos^2(x) - 1 = (sqrt(3)-1)sin(x)cos(x)

Thanks heaps

Hi

Did u mean solve the trigonometric equation ? If so ,

For (1) , $\displaystyle \frac{1-\tan^2 x}{1+\tan^2 x}+\cos x=0$

$\displaystyle \frac{1-\frac{\sin^2 x}{\cos^2 x}}{\sec^2 x}+\cos x=0$

$\displaystyle \frac{\cos^2 x-\sin^2 x}{\cos^2 x}(\cos^2 x)+\cos x=0$

$\displaystyle 2\cos^2 x+\cos x-1=0$

Then $\displaystyle (2\cos x-1)(\cos x +1)=0$ , i think you should be able to continue from here .
• Sep 21st 2009, 04:30 AM
deltaxray
Thanks Mate. (Happy)
• Sep 21st 2009, 04:46 AM
Grandad
Hello deltaxray

Welcome to Math Help Forum!
Quote:

Originally Posted by deltaxray
Can someone please help me solve the following trig equations please? Thanks

1-tan^2(x)/1+tan^2(x) + cos(x) = 0

(sqrt(3)+1)cos^2(x) - 1 = (sqrt(3)-1)sin(x)cos(x)

Thanks heaps

For question 2, divide both sides by $\displaystyle \cos^2x$:

$\displaystyle \sqrt3+1 - \sec^2x=(\sqrt3-1)\frac{\sin x\cos x}{\cos^2x}$

Then express everything in terms of $\displaystyle \tan x$. Re-arrange as a quadratic in $\displaystyle \tan x$, and you'll find it factorises into:

$\displaystyle (\tan x -1)(\tan x + \sqrt3)=0$

Can you fill in the gaps?

Grandad
• Sep 21st 2009, 05:26 AM
deltaxray
Thanks granddad. I got it (Nod)
• Sep 21st 2009, 05:50 AM
pacman
by inspection, pi is a root to the equation, the other 2 roots are found already; x = {pi/3, pi, 5pi/3}

the graph below confirms it