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

3 Answers
Apr 27, 2018

theta=pi/2, (3pi)/2, (7pi)/6, (11pi)/6θ=π2,3π2,7π6,11π6

Explanation:

We're going to want to get rid of the double angle, sin2theta.sin2θ.

Recall the identity sin2theta=2sinthetacosthetasin2θ=2sinθcosθ and apply it:

2sinthetacostheta+costheta=02sinθcosθ+cosθ=0

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

costheta(2sintheta+1)=0cosθ(2sinθ+1)=0

We now solve the following equations:
costheta=0cosθ=0
2sintheta+1=02sinθ+1=0

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

For costheta=0,cosθ=0, we see from the unit circle that this holds true for theta=pi/2, (3pi)/2θ=π2,3π2 over the interval [0,2pi).[0,2π).

2sintheta+1=02sinθ+1=0
2sintheta=-12sinθ=1
sintheta=-1/2sinθ=12

We see this holds true for theta=(7pi)/6, (11pi)/6θ=7π6,11π6 over [0, 2pi).[0,2π).

Apr 27, 2018

I tried this:

Explanation:

We can write:

sin2theta=2sinthetacosthetasin2θ=2sinθcosθ

so that our equation becomes:

2sinthetacostheta+costheta=02sinθcosθ+cosθ=0

collect:

costheta(2sintheta+1)=0cosθ(2sinθ+1)=0

so, this equation is satisfied either when:

costheta=0cosθ=0
i.e. when theta=pi/2 and theta=3/2piθ=π2andθ=32π

or when:
2sintheta+1=02sinθ+1=0
sintheta=-1/2sinθ=12
i.e. when theta=7/6pi and theta=11/6piθ=76πandθ=116π

Apr 27, 2018

theta in { pi/2, {7pi}/6, {11pi}/6, {3pi}/2 } θ{π2,7π6,11π6,3π2}

Explanation:

sin 2 theta + cos theta = 0sin2θ+cosθ=0

2 sin theta cos theta + cos theta = 02sinθcosθ+cosθ=0

cos theta ( 2 sin theta + 1) = 0cosθ(2sinθ+1)=0

cos theta = 0cosθ=0 or 2 sin theta + 1 = 02sinθ+1=0

cos theta = 0cosθ=0 or sin theta = - 1/2sinθ=12

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


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 ks.

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.