Find the sum of the three smallest positive values of θ such that 4\cos^2(2\theta-\pi) =3. (Give your answer in radians.)

Question

Find the sum of the three smallest positive values of θ such that
`4\cos^2(2\theta-\pi) =3`.

(Give your answer in radians.)

Answer 1

`(13)/(12)\pi`

Explanation:

`4\cos^2(2\theta-\pi)=3`

We want to first isolate the trig expression.

We will do this by first dividing both sides by 4 and then taking the square root of both sides.

`\cos^2(2\theta-\pi)=(3)/(4)`

`\cos(2\theta-\pi)=\pm \sqrt{(3)/(4)}`

Let's clean up the right hand side a little:

`\cos(2\theta-\pi)=\pm (√(3))/(2)`

Break time to think about something to help us solve the above:

So
`\cos(u)=(√(3))/(2)` when
`u=\pm (\pi)/(6)` in the one cycle of
`\cos(u)`.

So
`\cos(u)=-(√(3))/(2)` when
`u=\pm (5\pi)/(6)` in the one cycle of
`\cos(u)`.

Back to it:

`\cos(2\theta-\pi)=\pm (√(3))/(2)`

Solving one of those equations:

`\cos(2\theta-\pi)=(√(3))/(2)` when

`2\theta-\pi=\pm (\pi)/(6)+2\pi k`

Let's solve for
`\theta`.

Add
`\pi` on both sides:

`2\theta=\pm (\pi)/(6)+2\pi k+\pi`

Divide both sides by 2:

`\theta=\pm (\pi)/(12)+\pi k+(\pi)/(2)`

`\theta=\pm (\pi)/(12)+(\pi)/(2)+\pi k`

Now we also have to solve the other:

`\cos(2\theta-\pi)=-(√(3))/(2)` when

`2\theta-\pi=\pm (5\pi)/(6)+2\pi k`

Let's solve for
`\theta`.

Add
`\pi` on both sides:

`2\theta=\pm (5\pi)/(6)+2\pi k+\pi`

Divide both sides by 2:

`\theta=\pm (5\pi)/(12)+\pi k+(\pi)/(2)`

`\theta=\pm (5\pi)/(12)+(\pi)/(2)+\pi k`

So the full set of solutions is:

`\theta=\pm (\pi)/(12)+(\pi)/(2)+\pi k`

`\theta=\pm (5\pi)/(12)+(\pi)/(2)+\pi k`

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I'm going to evaluate these expressions for some integer
`k`:

`k=-2` we have:

`\theta=\pm (\pi)/(12)+(\pi)/(2)+\pi (-2)`

`\theta=(-17)/(12)\pi \text{ or } (-19)/(12)\pi`

`\theta=\pm (5\pi)/(12)+(\pi)/(2)+\pi (-2)`

`\theta=(-13)/(12)\pi \text{ or } (-23)/(12)\pi`

`k=-1` we have:

`\theta=\pm (\pi)/(12)+(\pi)/(2)+\pi (-1)`

`\theta=(-5)/(12)\pi \text{ or } (-7)/(12)\pi`

`\theta=\pm (5\pi)/(12)+(\pi)/(2)+\pi (-1)`

`\theta=(-1)/(12)\pi \text{ or } (-11)/(12)\pi`

`k=0` we have:

`\theta=\pm (\pi)/(12)+(\pi)/(2)+\pi (0)`

`\theta=(7)/(12)\pi \text{ or } (5)/(12)\pi`

`\theta=\pm (5\pi)/(12)+(\pi)/(2)+\pi (0)`

`\theta=(11)/(12)\pi \text{ or } (1)/(12)\pi`

`k=1` we have:

`\theta=\pm (\pi)/(12)+(\pi)/(2)+\pi (1)`

`\theta=(19)/(12)\pi \text{ or } (17)/(12)\pi`

`\theta=\pm (5\pi)/(12)+(\pi)/(2)+\pi (1)`

`\theta=(23)/(12)\pi \text{ or } (13)/(12)\pi`

Anyways the values will keep getting larger from here.

We can see that the smallest positive values happen when
`k=0`.

So we have the first smallest is: \frac{1}{12}\pi[/tex].

The second smallest is
`(5)/(12)\pi`.

The third smallest is
`(7)/(12)\pi`.

So the sum of these numbers are:

`(1)/(12)\pi+(5)/(12)\pi+(7)/(12)\pi`

`(1+5+7)/(12)\pi`

`(13)/(12)\pi`

Answer 2

∅= 13π/12

Explanation:

Solve until you get cos(___) on one side and a value on the other side.

I got cos(2∅-π) = ±√(3)/2

I know that the only cos() value that gives me ±√(3)/2 are -5π/6, -π/6, and π/6 from looking at the unit circle; those 3 are the smallest value that would give me ±√(3)/2.

So i solve for what value of 2∅-π would give me -5π/6, -π/6, and π/6

by solving for

2∅₁-π = -5π/6, 2∅₂-π = -π/6, and 2∅₃-π = π/6

I got ∅₁=π/12, ∅₂=5π/12, ∅₃=7π/12.

The question wanted the sum of the three smallest value so i added it up and got 13π/12.

And thanks for peer review. Good catch on my mistakes.