Question

How do you solve: sin2θ + cosθ = 0 between 0 and 2pi?

Answer 1

`theta=pi/2, (3pi)/2, (7pi)/6, (11pi)/6`

Explanation:

We're going to want to get rid of the double angle, `sin2theta.`

Recall the identity `sin2theta=2sinthetacostheta` and apply it:

`2sinthetacostheta+costheta=0`

Note that all of these terms involve `costheta,` so we can factor it out:

`costheta(2sintheta+1)=0`

We now solve the following equations:
`costheta=0`
`2sintheta+1=0`

Each coordinate pair `(x,y)` on this unit circle represents `(costheta,sintheta)` at an angle `theta.`

For `costheta=0,` we see from the unit circle that this holds true for `theta=pi/2, (3pi)/2` over the interval `[0,2pi).`

`2sintheta+1=0`
`2sintheta=-1`
`sintheta=-1/2`

We see this holds true for `theta=(7pi)/6, (11pi)/6` over `[0, 2pi).`

Answer 2

I tried this:

Explanation:

We can write:

`sin2theta=2sinthetacostheta`

so that our equation becomes:

`2sinthetacostheta+costheta=0`

collect:

`costheta(2sintheta+1)=0`

so, this equation is satisfied either when:

`costheta=0`
i.e. when `theta=pi/2 and theta=3/2pi`

or when:
`2sintheta+1=0`
`sintheta=-1/2`
i.e. when `theta=7/6pi and theta=11/6pi`

Answer 3

`theta in { pi/2, {7pi}/6, {11pi}/6, {3pi}/2 }`

Explanation:

`sin 2 theta + cos theta = 0`

`2 sin theta cos theta + cos theta = 0`

`cos theta ( 2 sin theta + 1) = 0`

`cos theta = 0` or `2 sin theta + 1 = 0`

`cos theta = 0` or `sin theta = - 1/2`

You knew a 30/60/90 or 45/45/90 triangle was going to show up sooner or later.

`theta = pi/2 + pi k quad or quad theta = -pi/6 + 2 pi k quad or quad theta = -{5pi}/6 + 2pi k`

The last two came from two supplementary angles (`-30^circ` and `-150^circ`) whose sines are `-1/2.`

Let's enumerate `0 \le theta \le 2pi`.

From `\pi/2 + \pi k` we get `theta = pi/2, {3pi}/2`

From `-pi/6 + 2 pi k` at `k=1` we get `theta={11pi}/6`

From `{-5pi}/6 + 2\pi k` at `k=1` we get `theta={7pi}/6`

Check: I'll check a couple, leave the rest for you.

`theta={3pi}/2 quad quad sin(3pi)+cos({3pi}/2) = 0+0 quad sqrt`

`theta={7pi}/6 quad quad sin({7pi}/3)+cos({7pi}/6) =sin(pi/3)-cos(pi/6)=0 quad sqrt`

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That's the end, but I'll keep writing so feel free to keep reading this optional part..

Getting those `pi k` and `2 pi k` right is a bit of an art. It helps to go back to first principles, which for me is `cos x = cos a` has solutions `x=\pm a + 2\pi k quad` integer `k`.

We rewrite `cos theta = 0` or `sin theta = - 1/2` as

`cos theta = cos(pi/2)` or `cos(pi/2 - theta) = cos({2pi}/3)`

Applying our solutions, `cos theta = cos(pi/2)` gives `theta = pm pi/2 + 2pi k`. That's really two lists, that can be combined giving `theta = pi/2 + pi k`. The same set of values is produced for different `k`s.

`cos(pi/2 - theta) = cos({2pi}/3)` gives

`pi/2 - theta = \pm {2pi}/3 +2pi k`

`theta = pi/2 \pm {2pi}/3 + 2 \pi k`

`theta = {7pi}/6 + 2\pi k` or `theta = -pi/6 + 2pi k`

which is essentially what we got.