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

3 Answers
Apr 27, 2018

#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#

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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).#

Apr 27, 2018

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#

Apr 27, 2018

#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#


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.