Question

Solving trig Equations
a.) sin^2x=3 cos^2x

Answer 1

Please Help!!!

Answer 2

reposting question in better format:
\[\sin^2(x) = 3\cos^2(x)\]

Answer 3

I do not have a clue where to even start.

Answer 4

well, as it stands it looks pretty bad... do you remember what sin(x)/cos(x) is equal to?

Answer 5

tan x

Answer 6

good! now, let's use that to make this question simpler. divide both sides by cos^2(x) and it starts looking better.

Answer 7

You should get this:\[\tan^2(x) = 3\]

Answer 8

So then you square root it right?

Answer 9

yes!

Answer 10

do not forget that you will have a positive and negative root

Answer 11

ok thank you

Answer 12

Let me know if you need any further help good luck!

Answer 13

Alternatively, you can use the identity
\[\sin^2x +\cos^2x=1\]
Rewriting the left (or right side, appropriately), you have
\[\sin^2x=3(1-\sin^2x)\\
\sin^2x=3-3\sin^2x\\
4\sin^2x=3\\
\sin^2x=\frac{3}{4}\]

Answer 14

Which gives you the difference of squares
\[\sin^2x-\left(\frac{\sqrt3}{2}\right)^2=0\\
\left(\sin x+\frac{\sqrt3}{2}\right)\left(\sin x-\frac{\sqrt3}{2}\right)=0\]
Easily solvable.