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 .
Hello deltaxray
Welcome to Math Help Forum!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