## anonymous 3 years ago Solving trig Equations a.) sin^2x=3 cos^2x

1. anonymous

2. jtvatsim

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

3. anonymous

I do not have a clue where to even start.

4. jtvatsim

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

5. anonymous

tan x

6. jtvatsim

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

7. jtvatsim

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

8. anonymous

So then you square root it right?

9. jtvatsim

yes!

10. jtvatsim

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

11. anonymous

ok thank you

12. jtvatsim

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

13. anonymous

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

14. anonymous

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.