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

Inna Shpiro | Certified Educator

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First, let us rewrite the cube of sine and cosine in the left-hand side of the above equation as a product of a square of the sine/cosine and the sine/cosine:

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

Now, group the product of sinx and sin(3x) together and apply the product sines formula. The same can be done with the product of cosx and cos(3x).

The general form of the formulas is as follows:

Product of sinesâ€”sina*sinb = 1/2(cos(a-b) - cos(a+b))

Product of cosinesâ€”cosa*cosb = 1/2(cos(a+b) + cos(a-b))

Accordingly, sinx*sin(3x) = 1/2(cos(2x) - cos(4x)), and cosx*cos(3x) = 1/2(cos2x + cos(4x)).

Plug these into the equation and multiply both sides by 2:

(sinx)^2*(cos(2x) - cos(4x)) + (cosx)^2*(cos(2x)+cos(4x)) = 0.

Now, open the parenthesis:

(sinx)^2*cos(2x) - (sinx)^2*cos(4x) + (cosx)^2*cos(2x) + (cosx)^2*cos(4x) = 0.

Group together the terms that contain cos(2x) and cos(4x):

cos(2x)*((sinx)^2 + (cosx)^2) + cos(4x)*((cosx)^2 - (sinx)^2) = 0.

Applying Pythagorean Theorem ((sinx)^2 +(cosx)^2 = 1) and the double-angle identity for cosine ((cosx)^2 - (sinx)^2 = 1) gives us the following:

cos(2x) +cos(4x)*cos(2x) = 0

Factoring out cos(2x) results in cos(2x)*(1 + cos4x) = 0.

We can now set each factor to zero and solve each of the resultant equations separately: cos(2x) = 0 and 1 + cos(4x) = 0.

Cosine is 0 when the angle is pi/2 or -pi/2 and for all coterminal angles:

2x = +-pi/2 + pi*k, where k is an integer (0, +-1, +-2, and so on).

Dividing this by 2 yields x = +-pi/4 +pi/2*k.

The second factor set to zero gives cos(4x) = -1. The solutions to this equation, however, are already included in the set above.

Therefore, the solution set of the above equation is x = +-pi/4 + pi/2*k, where k is an integer.

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giorgiana1976 | Student

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)}.