Question

How do you solve cos 2x − 2 sin2 x = 0 interval (−5, 5]?

Answer 1

`-256^@72; -166^@72; -76^@72; 13^@28; 103^@28; 193^@28`

Explanation:

cos 2x - 2sin 2 (1)
x = 0
Divide both sides by cos 2x.
Condition: `cos2x != 0`, --> `2x != pi/2`, and `2x != (3pi)/2`, -->
`x != pi/4`, and `x != (3pi)/4`.
Equation (1) becomes:
1 - 2 tan 2x = 0
`tan 2x = 1/2`
Calculator and unit circle give -->
`2x = 56^@56 + k180^@`.
`x = 13^@28 + k90^@`
The interval (-5, 5) is equivalent to interval `(-286^@62, 286^@62)`
k = 0 --> x = 13.28
k = -1 --> x = 13.28 - 90 = - 76.72 -->
k = -2 --> x = 13.28 - 180 = -16672
k = -3 --> x = 13.28 + 270 = -256.72
k = 1 --> x= 13.28 + 90 = 103.28
k = 2 --> x = 13.28 + 180 = 193.28
The answers for (-5, 5) are:
`-256^@72; -166^@72; -76^@72; 13^@28; 103^@28; 193^@28`

Answer 2

`x = "Arc""tan" ( 1/2(1-sqrt{5} ) ) pm 45^circ + 180^circ k quad` integer `k`

Explanation:

I'll take the questioner at their word and solve

`cos 2x - 2 sin 2x = 0`

Hopefully they didn't mean `sin^2 x` or `cos ^2 x`

There are a few different ways I can think of to solve this: (A) Let `y=2x` and invert `tan y`. (B) Use the double angle identities, then square to get a solvable quadratic for `cos^2 x.` (C) The linear combination of sine and cosine of the same angle is a scaling and a phase shift.

Let's try (C) first. We convert `(1,-2)` to polar coordinates:

`| (1","-2) |= sqrt{1^2+(-2)^2}=sqrt{5}`

This is fourth quadrant so we can use the principal value of the inverse tangent.

`alpha = angle(1,-2) = "Arc""tan"(-2) approx -63.43°`

We have

`1 = sqrt{5} cos alpha`

`-2 = sqrt{5} sin alpha`

`1 cos 2x - 2 sin 2x = 0`

`sqrt{5} cos alpha cos 2x + sqrt{5} sin alpha sin 2x = 0`

`sqrt{5} cos(2x - alpha) = 0`

`cos (2x - alpha) = cos (90^circ)`

`2x - alpha = pm 90^circ + 360^circ k`

`x = alpha/2 pm 45^circ + 180^circ k`

`x = 1/2 "Arc""tan"(-2) pm 45^circ + 180^circ k`

The factor of `1/2` on the arctangent is awkward.

We can make progress on

`theta = 1/ 2 alpha` where `alpha = angle(1,-2) = "Arc""tan"(-2)`

We know `alpha`'s in the fourth quadrant so

`cos alpha = 1/sqrt{5}`

`sin alpha = -2/sqrt{5}`

`tan (alpha/2) = sin alpha / {1 + cos alpha} = {1 - cos alpha}/sin alpha`

`tan(alpha/2) = {1 - 1/sqrt{5}}/(-2/sqrt{5}) = 1/2(1-sqrt{5})`

`alpha/2 = 1/2 "Arc""tan"(-2) = "Arc""tan" ( 1/2(1-sqrt{5} ) )`

`x = "Arc""tan" ( 1/2(1-sqrt{5} ) ) pm 45^circ + 180^circ k`

I gotta go so I'll let someone else convert to radians, pull out the answers in range and check it.

~~~~~~~~~~~~~

Attempts at other approaches:

Let `y=2x`

`cos y - 2sin y = 0`

`cos y = 2 sin y`

`tan y = sin y/ cos y= 1/2`

Invoking the multivalued arctangent,

`y = arctan (1/2)`

`2x = arctan(1/2)`

`x = 1/2 arctan (1/2)`

Similar to where we were but simpler.

~~~~~~~~~~~~~~~~~~

`2 cos^2 x - 1 - 2 cos x sin x = 0`

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

Let's square to get square of sines and cosines.

`(2 cos ^2 x-1)^2 = 4 cos^2 x sin ^2 x =4 cos ^2 x(1- cos ^2 x)`

Let `y = cos^2 x`

`(2y -1)^2 = 4y(1-y)`

`4y^2 - 4y + 1 = 4y - 4y^2`

`8y^2 - 8y+1 = 0`

`y = 1/8(4 pm sqrt{8}) = 1/4(2 pm sqrt{2})`

Uncle.