Trigonometry...how do i solve for x given: cos2x = cos^2 x, -pi < x < pi ???

Question

I dont understand the what to do with the range they give
- pi < x < pi ??????

Answer 1

To solve the above problem, use the double angle identity:

cos(2x) = cos^2(x) - sin^2(x)

Therefore, for

cos2x = cos^2(x)

Replace the left hand side with the identity

cos^2(x) - sin^2(x) = cos^2(x)

Subtract cos^2(x) both sides, to get

-sin^2(x) = 0
sin^2(x) = 0
sin(x) = 0

Where on that unit circle is sin(x) = 0? The answer to that is at 0 and pi.

However, -pi < x < pi means NOT INCLUDING pi.
Therefore, x = 0.

Answer 2

Sin2x is taken as sin(2x) in what follows: 2sinxcosx =3(sinx + 2(sinx)^2) grouping 1-cos2x, So sinx = 0 or 2cosx = 3(1+2sinx) So x = npi OR 4(cosx)^2 = 9 ( 1+4sinx + 4(sinx)^2) x = npi OR 4 - 4(sinx)^2 = 9 +36sinx + 36(sinx)^2 x = npi OR 40T^2 + 36T +5 = 0 [ T=sinx] x = npi OR T = [-9*rt(31)]/20 where * is + or -. Now use cal_C n get it

Answer 3

use formula cos2x = cos^2x -sin^2x
this gives
cos^2x-sin^2x=cos^2x
-sin^2x=0
==> sinx=0
==>x=0 degree or 0 radians,

note: 180 degree or pi radian is not an answer as the x is limited to -pi

Answer 4

That means that x is between the values of -pi and pi. Since, cosine is a periodic function (it repeats itself over and over) there are an inifinite number of values x could have been if they didn't restrict it.