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Solve equation cos 3xcos^3 (x)+sin3x sin^3(x)=0?

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maxxam10 | Student, College Freshman | eNoter

Posted May 12, 2013 at 5:11 PM via web

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Solve equation cos 3xcos^3 (x)+sin3x sin^3(x)=0?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 12, 2013 at 5:51 PM (Answer #1)

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You need to use the triple angle formula, such that:

`sin 3x = 3 sin x - 4sin^3 x`

`cos 3x = 4cos^3 x - 3 cos x`

Replacing `4cos^3 x - 3 cos x` for `cos 3x` and `3 sin x - 4sin^3 x ` for `sin 3x` , yields:

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

`4cos^6 x - 3cos^4 x + 3sin^4 x - 4sin^6 x = 0`

You need to group the terms, such that:

`(4cos^6 x - 4sin^6 x) - (3cos^4 x - 3sin^4 x) = 0`

`4(cos^6 x - sin^6 x) - 3(cos^4 x - sin^4 x) = 0`

`4(cos^6 x - sin^6 x) - 3(cos^4 x - sin^4 x) = 0`

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

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

Factoring out `(cos x - sin x)(cos x + sin x)` yields:

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

Using the zero product rule yields:

`cos x - sin x = 0 => cot x - 1 = 0 => cot x = 1 => x = pi/4 + npi`

`cos x + sin x = 1 => cot x = -1 => x = -pi/4 + npi`

`4 - 4sin^2 x*cos^2 x - 3 = 0 => 1 - 4sin^2 x*cos^2 x = 0`

`sin^2 x*cos^2 x = 1/4` invalid since only `sin x = cos x = sqrt2/2` at `x = pi/4`

Hence, evaluating the solution to the given trigonometric equation yields `x = +-pi/4 + npi.`

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