So I have two angle problems...

sin2x = 2cosxcos2x ; which becomes

0 = 2cosxcos2x - sin2x

0 = 2cosxcos2x - 2sinxcosx

0 = 2cosx(cos2x - sin x)

so either

2cosx = 0 OR cos2x - sinx = 0

Thank you.

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- Jan 2nd 2006, 05:02 PMCrimesofParisAngle Question
So I have two angle problems...

sin2x = 2cosxcos2x ; which becomes

0 = 2cosxcos2x - sin2x

0 = 2cosxcos2x - 2sinxcosx

0 = 2cosx(cos2x - sin x)

so either

2cosx = 0 OR cos2x - sinx = 0

Thank you. - Jan 2nd 2006, 06:30 PMThePerfectHackerQuote:

Originally Posted by**CrimesofParis**

Okay you did everything correctly up till now. Now you are required to solved for .

The first one . Divide by two,

, now you need to know when is zero. That is when as two non-coterminal angels. To find all solutions it is the same as and where is any integer. Because as you know that addition and subtraction of 360 keeps the trigonometric function of that angel the same. But is also an integer. Thus, all solutions for have form. or in radian form - Jan 2nd 2006, 06:47 PMThePerfectHacker
For your second question:

Use the identity

Thus,

Change signs,

Factor,

Now each one is set to zero,

Thus,

That is when as the only two non-coterminal angels. Thus all solution have form But,

Thus, all solutions have form or in radian form - Jan 2nd 2006, 06:55 PMThePerfectHacker
Finally the last factor equal to zero, (of the quadratic)

Thus,

That happens when and that is the only non-coterminal angle. Thus all solutions have form . But .

Thus, all solutions are . Or in radian form .

If you have difficulty understanding all that manipulating after the solutions just understand that by adding 360 or subtracting 360 you change nothing. Thus, those are**all**the solutions. All I did was manipulated into a more elegant form. - Jan 2nd 2006, 07:36 PMCrimesofParis
Thank you so much, that pretty much answers my questions, except I forgot to mention the answer has to be radians, so I should be able to covert the middle steps to radians. But thank you.