OpenStudy (anonymous):

Solve for x: 3sinx^2=cosx^2

3 years ago
OpenStudy (dls):

Divide both sides by cos^2 x to get \[\LARGE 3 \tan^2x = 1\] \[\LARGE \tan^2 x= \frac{1}{3} = > \tan x = \pm \frac{1}{3}\] now you can solve for x

3 years ago
OpenStudy (dls):

I hope you meant cos^2x and not cosx^2

3 years ago
OpenStudy (anonymous):

yes that is what I meant. \[3\sin ^{2}x=\cos ^{2}x\] \[0\le x <2\pi \]

3 years ago
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