Find the exact solutions of the equations in the given interval.

Question

sin(3θ) = − 1/2 in (−∞, ∞)

cos(5x) = √3 / 2 in [0, 2π)

cos(4θ) = 1/2 in [0, π)

cos (2x) = sin(x) in [0, 2π)

Answer 1

First part:

sin (3θ) = -1/2

In the unit circle we see that the angle whose sine is -1/2 occurs in two places. 7π/6 and 11π/6

So. 3θ = 7π/6 and 3θ = 11π/6. dividing both sides by 3 gives.

θ = 5π/18 and θ = 11π/18. This describes the values within one revolution between 0 and 2π but

we are asked for all the solutions between -∞ and ∞ .

So the solutions are 5π/18 plus or minus 2π. and 11π/18 plus or minus 2π.

Second part:

cos(5x) = √3 / 2 in [0, 2π)

Again from the unit circle we know that the angle whose cosine is √3 /2. occurs in two places. At π/6 and at 11π/6. These values are the only ones in the interval [0, 2π) .

So 5x = π/6. and so dividing by 5 gives x = π/30. And 5x = 11π/6 then dividing by 5 gives x= 11π/30.

Third part:

cos(4θ) = 1/2 in [0, π)

Again, in the unit circle the angle whose cosine is 1/2 occurs in two places. At π/3 and at 5π/3.

So 4θ = π/3. Dividing both sides by 4 gives θ = π/12.

And 4θ = 5π/3. Dividing both sides by 4 gives θ = 5π/12

Both of these answers appear in the interval [0, π)

Fourth part:

cos (2x) = sin(x) in [0, 2π)

Using the double angle formula for cos (2x) = 1 - 2sin²x

Then 1 - 2sin²x = sin x

Setting this equation equal to 0 gives. 0 = 2sin²x + sin x - 1

Factoring. 0 = (2 sin x - 1) (sin x + 1)

Then 0 = 2 sin x - 1. and 0 = sin x + 1

So sin x = 1/2. and sin x = -1

Unit circle x = π/6 and 5π/6. and x = 3π/2

These 3 solutions fall within the interval [0, 2π).

The unit circle is incredibly useful.

Answer 2

sin 3θ = -1/2

3θ = 11π / 6

θ = 11π / 18

Others done in same manner.