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Trig question

Hey guys,

I was wondering if you could explain this question to me
solve the equation cos2x= the squarerootof 3 sin2x

if i bring sin 2x under cos2x to get tan2x

so the new equation would be tan2x= squareroot of 3 would it not?
Reply 1
Original post by HEY_101
Hey guys,

I was wondering if you could explain this question to me
solve the equation cos2x= the squarerootof 3 sin2x

if i bring sin 2x under cos2x to get tan2x

so the new equation would be tan2x= squareroot of 3 would it not?


cos2x=3sin2x \displaystyle cos2x = \sqrt{3sin2x}

Square both sides, to get, cos22x=3sin2x cos^22x=3sin2x

Remember, cos22x=1sin22x cos^22x = 1-sin^22x

This will give you a quadratic.
Reply 2
Unless

cos2x=3sin2xcos2x = \sqrt3 \sin2x

in which case

No

cosxsinxtanx\dfrac{cosx}{sinx} \not= tanx
Reply 3
Original post by raheem94
cos2x=3sin2x \displaystyle cos2x = \sqrt{3sin2x}


the square root is only on the 3 not the sin 2x
Reply 4
Original post by HEY_101
I was thinking that as well but the answers says tan2x= 1/ square root of 3.


cos2x=3sin2x \displaystyle cos2x = \sqrt3 sin2x

Divide both sides by cos2x cos2x
cos2xcos2x=3sin2xcos2x    1=3tan2x    tan2x=13 \displaystyle \frac{cos2x}{cos2x} = \frac{\sqrt3 sin2x}{cos2x} \implies 1 = \sqrt3 tan2x \implies tan2x = \frac{1}{\sqrt3}

I don't understand why are you getting confused at such a simple thing.

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