# Trigonometric equation

• September 21st 2009, 03:18 AM
deltaxray
Trigonometric equation

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
• September 21st 2009, 04:39 AM
Quote:

Originally Posted by deltaxray

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

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

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

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

Then $(2\cos x-1)(\cos x +1)=0$ , i think you should be able to continue from here .
• September 21st 2009, 05:30 AM
deltaxray
Thanks Mate. (Happy)
• September 21st 2009, 05:46 AM
Hello deltaxray

Welcome to Math Help Forum!
Quote:

Originally Posted by deltaxray

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 $\cos^2x$:

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

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

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

Can you fill in the gaps?

• September 21st 2009, 06:26 AM
deltaxray
Thanks granddad. I got it (Nod)
• September 21st 2009, 06:50 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