trigonometry ecuationFind the solutions of the equation: cos x + cos 3x = 0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may use triple angle formula for `cos 3x` , such that:

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

Substituting `4cos^3 x - 3 cos x` for `cos 3x ` yields:

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

`4cos^3 x - 2 cos x = 0`

Factoring out `2 cos x` yields:

`2 cos x(2cos^2 x - 1) = 0 => {(2cos x = 0),(2cos^2 x - 1 = 0):}`

`{(cos x = 0),(cos^2 x = 1/2):} => {(x = +-pi/2 + 2n*pi),(cos x = +-sqrt(1/2)):}`

`cos x = sqrt2/2 => x = +-pi/4 + 2n*pi`

`cos x = -sqrt2/2 => x = +-(pi - pi/4) + 2n*pi => x = +-(3pi)/4 + 2n*pi`

Hence, evaluating the solutions to the given equation, using the triple angle identity, yields `x = +-pi/4 + 2n*pi, x = +-pi/2 + 2n*pi, x = +-(3pi)/4 + 2n*pi.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll transform the sum in product using the formula:

cos a + cos b = 2 cos[(a+b)/2]cos [(a-b)/2]

We'll write the formula for a = x and b = 3x:

cos x + cos 3x = 2cos[(x+3x)/2]cos [(x-3x)/2]

cos x + cos 3x = 2cos[(4x)/2]cos [(-2x)/2]

cos x + cos 3x = 2cos[(2x)]cos [(-x)]

Since the function cosine is even, we'll write cos [(-x)] = cos x.

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

We'll put 2cos 2x*cos x = 0

cos 2x = 0

2x = +/- arccos 0 + 2kpi

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

x = +/- pi/4 + kpi

cos x = 0

x = +/- pi/2 + 2kpi

The solutions of the equation are: {+/- pi/4 + kpi}U{+/- pi/2 + 2kpi}.

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