# I need to find the solution of equation sin^3x*sin3x+cos^3x*cos3x=0

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll use the identity of triple angle:

(sin x)^3 = (3sin x - sin 3x)/4

(cos x)^3 = (cos 3x + 3cos x)/4

We'll substitute th reltions above into equation:

sin 3x*(3sin x - sin 3x)/4 + cos 3x*(cos 3x + 3cos x)/4 = 0

We'll remove the brackets:

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

We'll group the 1st and the last terms:

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

We notice the formula:

cosx*cos 3x + sin x*sin 3x = cos(3x - x) = cos 2x

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

3cos 2x + cos 6x = 0

3cos 2x + cos 3*2x = 0

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

We'll eliminate like terms:

4(cos 2x)^3 = 0

We'll divide by 4:

(cos 2x)^3 = 0 <=> cos 2x = 0

2x = +/-arccos 0 + 2kpi

2x = +/-(pi/2) + 2kpi

x = +/-(pi/4) + kpi

x = (2k + 1)*(pi/4), where k is an integer number.

The solutions of the equation are: {(2k + 1)*(pi/4)}.