Question

How do I solve `(2sqrt2-2)cosalpha=3sin^2alpha+sqrt2-4`?

Answer 1

`:. alpha=2kpi+-arccos[{-(sqrt2-1)+-root(4)2}/3], k in ZZ`.

Explanation:

Replacing `sin^2alpha` in the right member by `(1-cos^2alpha)`, we have,

`(2sqrt2-2)cosalpha=3-3cos^2alpha+sqrt2-4`.

`:. 3cos^2alpha+2(sqrt2-1)cosalpha-(sqrt2-1)=0`.

This is a quadr. eqn. in `cosalpha`, and, its discriminant

`Delta=(2sqrt2-2)^2+4*3*(sqrt2-1)`,

`=(8-8sqrt2+4)+12sqrt2-12`,

`:. Delta=4sqrt2`.

Therefore, using the quadr. formula, we have,

`cosalpha={-2(sqrt2-1)+-sqrt(4sqrt2)}/(2*3)`,

`={-2(sqrt2-1)+-2root(4)2}/(2*3)`,

`:. cosalpha={-(sqrt2-1)+-root(4)2}/3`.

`:. alpha=2kpi+-arccos[{-(sqrt2-1)+-root(4)2}/3], k in ZZ`.

Answer 2

`=>alpha=2npi+-cos^-1((1-sqrt2+2^(1/4))/3)` or
`alpha=2npi+-cos^-1((1-sqrt2-2^(1/4))/3)`

Explanation:

`" "(2sqrt2-2)cosalpha=3sin^2alpha+sqrt2-4`
`=>2(sqrt2-1)cosalpha=3(1-cos^2alpha)+sqrt2-4`

let `cosalpha=y", then equation becomes"`
`2(sqrt2-1)y=3(1-y^2)+sqrt2-4`
`=>3y^2+2(sqrt2-1)y-3-sqrt2+4=0`
`=>3y^2+2(sqrt2-1)y+(1-sqrt2)=0`

now solve for `y`, from quadratic formula
`y=(-2(sqrt2-1)+-sqrt((2(sqrt2-1))^2-4(3)(1-sqrt2)))/(2(3))`
`=(2(1-sqrt2)+-sqrt(4(2+1-2sqrt2)-12+12sqrt2))/6`
`=(2-2sqrt2+-2sqrt(sqrt2))/6`
`=>cosalpha=(2-2sqrt2+-2sqrt(sqrt2))/6`
`=> cosalpha=(1-sqrt2+2^(1/4))/3" or " cosalpha=(1-sqrt2-2^(1/4))/3`
`=>alpha=cos^-1((1-sqrt2+2^(1/4))/3)` or
`alpha=cos^-1((1-sqrt2-2^(1/4))/3)`
since `cosx` repeats itself after `2npi` period `(ninZZ) " add " 2npi` for general solutions

Answer 3

`x = +- 74^@93 + k360^@`
`x = +- 122^@68 + k360^@`

Explanation:

`(2sqrt2 - 2)cos x = 3sin^2 x + sqrt2 - 4`
`(2sqrt2 - 2)cos x = 3 - 3cos^2 x + sqrt2 - 4`
`3cos^2 x + 2(sqr2 - 1))cos x - (sqrt2 - 1) = 0`
`3cos^2 x + 0.828cos x - 0.414 = 0`
Solve this quadratic equation for cos x.
`D = d^2 - 4ac = 0.686 + 4.968 = 5.65` --> `d = +- 2.38`
There are 2 real roots:
`cos x = -b/(2a) +- d/(2a) = - 0.828/6 +- 2.38/6 = - 0.138 +- 0.40`
cos x = - 0.54, and cos x = 0.26
a. cos x = - 0.54
Calculator and unit circle give 2 solutions:
`x = +- 122^@68 + k360^@`
b. cos x = 0.26 -->
`x = +- 74^@93 + k360`
Check by calculator:
1. x = 74.93 --> 0.83cos x = 0.22 --> `(2sqr2 - 2)cos x = 0.22`
`3sin^2 x + sqrt2 - 4 = 2.8 + 1,41 - 4 = 0.21`. Proved
2. x = 122.68 --> `(2sqrt2 - 2)cos x = - 0.45`
`3sin^2 x + sqrt2 - 4 = 2.13 + 1.42 - 4 = - 0.45`. Proved