Question

Find the solutions of cos^2 x - sin^2 x - cos6x = 0 in the interval [0, 2pi].

Answer 1

A friendly advice:
\[\cos^2x-\sin^2x-\cos6x=0\]
You can always use the fundamental identity to write \(\sin^2x\) in terms of cosine, which is equal to \(\sin^2=1-\cos^2x\):
\[\cos^2x-\sin^x-\cos6x=0 \iff \cos^2x-(1-\cos^2x)-\cos6x=0 \]
\[\cos^2x-1+\cos^2x-\cos6x=0 \iff 2\cos^2x-\cos6x-1=0 \]
And so on.

Answer 2

That's where I'm getting stuck. The -cos6x seems to make things difficult.

Answer 3

cos^2 x - sin^2 x - cos(6x) = 0
cos(2x) - cos(6x) = 0
cos(2x) - cos(2x+4x) = 0
cos(2x) - cos(2x)cos(4x) + sin(2x)sin(4x) = 0
cos(2x) - cos(2x)(cos^2(2x)-sin^2(2x)) + 2sin(2x)sin(2x)cos(2x) = 0
cos(2x) - cos^3(2x) + cos(2x)sin^2(2x) + 2sin^2(2x)cos(2x) = 0
cos(2x) - cos^3(2x) + cos(2x)(1-cos^2(2x)) + 2sin^2(2x)cos(2x) = 0
cos(2x) - cos^3(2x) + cos(2x) - cos^3(2x) + 2(1-cos^2(2x))cos(2x) = 0
cos(2x) - cos^3(2x) + cos(2x) - cos^3(2x) + 2cos(2x) - 2cos^3(2x) = 0
-4cos^3(2x) + 4cos(2x) = 0

Answer 4

4a^3 - 4a = 0
4a(a^2 - 4) = 0
4a(a+2)(a-2) = 0
a=0,2,-2

cos(2x) \(\ne\) \(\pm\)2
because cosine is always between -1 and 1.

cos(2x)=0
and the solution is a general solution to this equation

Answer 5

I don't see any mistakes but for some reason that equation is not giving all solutions to the initial equation.

Answer 6

oh nevermind I see a mistake

Answer 7

\[4a^3-4a=0 \\ 4a^2(a-1)=0\]

Answer 8

awesome trig work @fbi2015