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haleyking345
Solving trig Equations a.) sin^2x=3 cos^2x
reposting question in better format: \[\sin^2(x) = 3\cos^2(x)\]
I do not have a clue where to even start.
well, as it stands it looks pretty bad... do you remember what sin(x)/cos(x) is equal to?
good! now, let's use that to make this question simpler. divide both sides by cos^2(x) and it starts looking better.
You should get this:\[\tan^2(x) = 3\]
So then you square root it right?
do not forget that you will have a positive and negative root
Let me know if you need any further help good luck!
Alternatively, you can use the identity \[\sin^2x +\cos^2x=1\] Rewriting the left (or right side, appropriately), you have \[\sin^2x=3(1-\sin^2x)\\ \sin^2x=3-3\sin^2x\\ 4\sin^2x=3\\ \sin^2x=\frac{3}{4}\]
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.