Thread: Angle 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.

Originally Posted by CrimesofParis

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.

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

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

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.

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.