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
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
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.