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haleyking345
 3 years ago
Solving trig Equations
a.) sin^2x=3 cos^2x
haleyking345
 3 years ago
Solving trig Equations a.) sin^2x=3 cos^2x

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jtvatsim
 3 years ago
Best ResponseYou've already chosen the best response.1reposting question in better format: \[\sin^2(x) = 3\cos^2(x)\]

haleyking345
 3 years ago
Best ResponseYou've already chosen the best response.0I do not have a clue where to even start.

jtvatsim
 3 years ago
Best ResponseYou've already chosen the best response.1well, as it stands it looks pretty bad... do you remember what sin(x)/cos(x) is equal to?

jtvatsim
 3 years ago
Best ResponseYou've already chosen the best response.1good! now, let's use that to make this question simpler. divide both sides by cos^2(x) and it starts looking better.

jtvatsim
 3 years ago
Best ResponseYou've already chosen the best response.1You should get this:\[\tan^2(x) = 3\]

haleyking345
 3 years ago
Best ResponseYou've already chosen the best response.0So then you square root it right?

jtvatsim
 3 years ago
Best ResponseYou've already chosen the best response.1do not forget that you will have a positive and negative root

jtvatsim
 3 years ago
Best ResponseYou've already chosen the best response.1Let me know if you need any further help good luck!

SithsAndGiggles
 3 years ago
Best ResponseYou've already chosen the best response.0Alternatively, you can use the identity \[\sin^2x +\cos^2x=1\] Rewriting the left (or right side, appropriately), you have \[\sin^2x=3(1\sin^2x)\\ \sin^2x=33\sin^2x\\ 4\sin^2x=3\\ \sin^2x=\frac{3}{4}\]

SithsAndGiggles
 3 years ago
Best ResponseYou've already chosen the best response.0Which 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.
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