Solve the following equation in the interval [0, 2pi]. 2(sin(t))^2-sin(t)-1=0?

Question

I'm assuming I can change sin(x) to any replacement variable temporarily. Making sin(x)=a I get stuck at the part where I need to use the quadratic formula. 2a^2-a-1=0 --> (2a+1)(a-1)=0 --> sin(t)=-1/2 and sin(t)=1. Aren't my answers supposed to be 7pi/6, 11pi/6, and pi/2? The online homework thing I do this on isn't taking that so I assume my math is wrong. Where did I mess up and what am I supposed to do?

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Answer 1 · 2018-05-09T15:37:50

The questioner is totally correct. Not sure why their online submission didn't work; those things bite.

#t = {7pi}/6 or {11pi}/6 or pi/2#

The description is correct and would make a fine answer.

# 2 sin^2 t - sin t - 1 = 0 #

We could substitute #x=sin t# for a "regular" quadratic equation, but it's easy enough to factor it just like this:

# ( 2 sin t + 1 )(sin t - 1) = 0#

#sin t = -1/2 or sin t = 1 quad # just like the asker said

Of course the former is trig's biggest cliche, 30/60/90, this time in the third and forth quadrant. So from #sin t = -1/2 # we get

#t = 210^circ = {7pi}/6 or t=330^circ={11pi}/6#

and from #sin t = 1#

#t = 90^circ = pi/2#

The questioner is totally correct.

Check:

# 2 sin ^2(pi/2) - sin(pi/2)-1=2(1^2)-1-1=0 quad sqrt#

# 2 sin^2 ({7pi}/6) - sin ({7pi}/6)-1= 2(-1/2)^2-(-1/2)-1=0 quad sqrt#

#{11 pi}/6 # works too, as you may wish to verify.