# Solving with Trig Identities..

• Dec 3rd 2009, 09:48 AM
Savior_Self
Solving with Trig Identities..
Solve the following equations given that $\displaystyle 0 \leq x \leq 2\pi$.

$\displaystyle a) \cos^2x - \sin^2x = 1$

$\displaystyle b) 1 + \sqrt{3}\tan2x = 0$

Your help is really appreciated. Thanks.
• Dec 3rd 2009, 10:09 AM
Light
$\displaystyle a) \cos^2x - \sin^2x = 1$

$\displaystyle b) 1 + \sqrt{3}\tan2x = 0$

Your help is really appreciated. Thanks.[/quote]
a) Replace sin^2 X by 1-cos^2 X.
The equation becomes 2cos^2 X =2
• Dec 3rd 2009, 10:14 AM
Light
Quote:

Originally Posted by Savior_Self
Solve the following equations given that $\displaystyle 0 \leq x \leq 2\pi$.

$\displaystyle a) \cos^2x - \sin^2x = 1$

$\displaystyle b) 1 + \sqrt{3}\tan2x = 0$

Your help is really appreciated. Thanks.

b) tan2X = -1/root3
Then set the range according to 2X. that is From 0<2X<4pi
then replace 2X by another variable say Y
tan Y= -1/root3
solve for Y in the new given range.
At the end divide the answers by 2, because 2X=Y. Is that Ok.