Question

Question #85768

Answer 1

`{22.5˚, 202.5˚}`

Explanation:

We have that `cos2x = 2cos^2x - 1` and `sin2x = 2sinxcosx`.

`cos2x = sin2x + 1 - 1`

`cos2x = sin2x`

`cos2x- sin2x= 0`

We now let `u = 2x`.

`cosu = sinu`

Now consider the `45˚-45˚-90˚` triangle.

Notice that `sin(45˚) = cos(45˚) = 1/sqrt(2)`. This will also be the case when tangent only is positive in the third quadrant. This is because
the 3rd quadrant is the only quadrant where both sine and cosine are negative, just like the 1st quadrant is the only quadrant where both sine and cosine are positive.

We therefore have that `u = 45˚` and `u = 225˚`. Now we have to reverse the substitution. If our original equation was in `x`, our solution must also be in `x`.

`u = 2x`

For `u = 45˚`

`45 = 2x`

`x = 22.5˚`

For `u = 225˚`

`225 = 2x`

`x = 112.5˚`

However, if we check in the initial equation, we get that `x = 112.5˚` is not a valid solution.

The periodicity of `y = sin2x` is `360/2 = 180`. Therefore, we will be able to fit in another angle in this interval.

Our solution set in `0 ≤ x ≤ 2pi` is therefore `{22.5˚, 202.5˚}`.

**Practice exercises **

1. Solve the following equation for `x` in the interval `0 ≤ x < 2pi`. Note: the identity `sin^2x + cos^2x = 1` may be useful.

`(cosx - sinx)(cosx + sinx) + 2 = 2sinxcosx + 3`

Solution

`{0˚, 45˚, 135˚, 180˚, 225˚, 315˚}`

Hopefully this helps, and good luck!

Answer 2

22.5; 202.5
112.5; 292.5

Explanation:

`2cos^2 x - 1 = sin 2x`
`cos 2x = sin 2x`
`sin 2x - cos 2x = 0`
Use trig identity:
`sin a - cos a = - sqrt2cos (a + pi/4)`
In this case;
`sin 2x - cos 2x = - sqrt2cos (2x + pi/4) = 0`
`cos (2x + pi/4) = 0`
Unit circle gives 2 solutions:
a. `(2x + 45) = 90` --> `2x = 90 - 45 = 45 + k360`
`x = 22.5 + k180`
b. `2x + 45 = 270` --> `2x = 270 - 45 = 225 + k360`
`x = 112.5 + k180`
Answers for (0, 360);
22.5; 202.5
112.5; 292.5