Solving with Trig Identities..

**Savior_Self** · December 3rd 2009, 09:48 AM

Solve the following equations given that $0 \leq x \leq 2\pi$ .

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

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

Your help is really appreciated. Thanks.

**Light** · December 3rd 2009, 10:09 AM

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

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

Your help is really appreciated. Thanks.[/quote]
a) Replace sin^2 X by 1-cos^2 X.
The equation becomes 2cos^2 X =2

**Light** · December 3rd 2009, 10:14 AM

Originally Posted by Savior_Self

Solve the following equations given that $0 \leq x \leq 2\pi$ .

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

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

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.

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