50 day-streak problem

Geometry Level 4

how many $x \in {[{-\pi},\pi]}$ such that

$\cos 3x + \cos x = 0$

2 solutions

Prakhar Bindal
Aug 31, 2016

cos3x = 4cos^3(x) - 3cosx

so 4cos^3(x) = 2cosx

hence cosx = 0 or cos^2(x) = 1/2

the former has two solution viz pi/2 and -pi/2.

cos^(2)x is periodic with pi .

Two solutions in (0,pi) hence total of 4 solutions.

Finally adding a grand of 6 solutions

Vineet Golcha
May 21, 2015

cos(3x) = cos(2x + x), and simplify using the sum of cosines formula: cos (a+b) = cos(a) cos(b) - sin(a) sin(b)

cos(3x) + cos(x) = 0, using the above formula you get:

cos(2x) cos(x) - sin(2x) sin(x) +cos(x) = 0

changing cos(2x) to cos(x + x) and sin(2x) to sin(x + x), which become cos(2x) = (cos(x))^2 - (sin(x))^2, and sin(2x) = 2 sin(x) cos(x) :

[(cos(x))^2 - (sin(x))^2] cos(x) - [2 sin(x) cos(x)] sin(x) + cos(x) =0

(cos(x))^3 - (sin(x))^2 cos(x) - 2 (sin(x))^2*cos(x) + cos(x) = 0

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

change (sin(x))^2 to (1-(cos(x))^2):

(cos(x))^3 - 3 [1 - (cos(x))^2] cos(x) + cos(x) = 0

(cos(x))^3 - 3*[cos(x) - (cos(x))^3] + cos(x) =0

4 (cos(x))^3 - 2 cos(x) = 0 [Eqn.1]

[4 (cos(x))^2 - 2] cos(x) = 0 therefore (cos(x))=0 x=90,270degrees

(cos(x))^2 = 0.5

cos(x) = +/-(1/2)^0.5

or x = +/- pi/4 (45 degrees or 135 or 225 or 315) TOTAL=6 values