# 11/6(1+cosx)=5sin^2x+2cos2x x? solve

Asked on by sananab

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should write the equation in terms of `cos x`  only, hence, you should use the fundamental formula of trigonometry and the double angle formula such that:

`sin^2 x = 1 - cos^2 x`

`cos 2x = 2cos^2 x - 1`

Hence, substituting `1 - cos^2 x`  for `sin^2 x`  and `2cos^2 x - 1`  for `cos 2x`  yields:

`(11/6)(1+cosx) = 5(1 - cos^2 x) + 2(2cos^2 x - 1)`

You need to open the brackets such that:

`11/6 + 11/6cos x = 5 - 5cos^2 x + 4cos^2 x - 2`

`11/6 + 11/6cos x = 3 - cos^2 x `

You need to bring all terms to the left side such that:

`cos^2 x - 11/6cos x + 11/6 - 3 = 0`

`6cos^2 x - 11cos x -7 = 0`

You should come up with the substitution cos x = y such that:

`6y^2 - 11y -7 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (11 +- sqrt(121 + 168))/12`

`y_(1,2) = (11 +- sqrt289)/12`

`y_(1,2) = (11 +- 17)/12 =gt y_1 = 28/12 =gt y_1 = 7/3`

`y_2 = -6/12 =gt y_2 = -1/2`

You should solve for x the following equations `cos x = y_1`  and `cos x = y_2`  such that:

`cos x = 7/3`  is impossible to be solved since the values of cosine function are not larger than 1.

`cos x = -1/2 =gt x = +-cos^(-1)(-1/2) + 2npi`

`x = +-(pi - pi/3) + 2npi =gt x = +-2pi/3 + 2npi`

Hence, evaluating the general solution to the given equation yields  `x = +-2pi/3 + 2npi` .

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