Solveeit Logo

Question

Question: How do you solve \(\sin 2\theta + \cos \theta = 0\) where, \(0 < \theta < 2\pi \)?...

How do you solve sin2θ+cosθ=0\sin 2\theta + \cos \theta = 0 where, 0<θ<2π0 < \theta < 2\pi ?

Explanation

Solution

In this question, we are given a trigonometric equation and the limit between which θ\theta lies. Start by expanding the first term of the given equation. Take a term common and this will give you two factors. Keep each factor equal to 00. Then, find the value or values of θ\theta . Use the given limit to find the required values.

Formula used: sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta

Complete step-by-step solution:
The trigonometric equation given to us is sin2θ+cosθ=0\sin 2\theta + \cos \theta = 0.
Firstly, expand the first term of the given equation using a basic trigonometric identity.
sin2θ+cosθ=0\Rightarrow \sin 2\theta + \cos \theta = 0
Using sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta ,
2sinθ.cosθ+cosθ=0\Rightarrow 2\sin \theta .\cos \theta + \cos \theta = 0
Taking cosθ\cos \theta common,
cosθ(2sinθ+1)=0\Rightarrow \cos \theta \left( {2\sin \theta + 1} \right) = 0
Now, we have two factors. We will keep each factor equal to 00.
cosθ=0,2sinθ+1=0\Rightarrow \cos \theta = 0,2\sin \theta + 1 = 0
Next step is to find those angles at which the equation holds true.
First, we will solve the first part. We know that cosθ\cos \theta is equal to 00 at π2\dfrac{\pi }{2} and 3π2(=π+π2)\dfrac{{3\pi }}{2}\left( { = \pi + \dfrac{\pi }{2}} \right).
Hence, θ=π2,3π2\theta = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}.
Now, we will solve the second part.
2sinθ+1=0\Rightarrow 2\sin \theta + 1 = 0
sinθ=12\Rightarrow \sin \theta = \dfrac{{ - 1}}{2}
We know that sinθ\sin \theta is negative in the third and fourth quadrant. Therefore, our required values will also lie in those quadrants only. We know that sinθ\sin \theta is equal to 12\dfrac{1}{2} at π6\dfrac{\pi }{6} .
Therefore, its negative will be on 7π6(=π+π6)\dfrac{{7\pi }}{6}\left( { = \pi + \dfrac{\pi }{6}} \right) and 11π6(=2ππ6)\dfrac{{11\pi }}{6}\left( { = 2\pi - \dfrac{\pi }{6}} \right).

Hence, θ=π2,3π2,7π6,11π6\theta = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}.

Note: What if the range within with θ\theta lies was not given? Then, till what value would you have found the answer? In such situations, we find a general solution. Each trigonometric ratio has its different general solution. Let us see what would have been our answer, if there had been no range.
cosθ=0,sinθ=12\Rightarrow \cos \theta = 0,\sin \theta = \dfrac{{ - 1}}{2}
We can write it as –
cosθ=cosπ2,sinθ=sin7π6\Rightarrow \cos \theta = \cos \dfrac{\pi }{2},\sin \theta = \sin \dfrac{{7\pi }}{6}
Putting them in their general formula –
For cosθ\cos \theta ,
If cosθ=cosα\cos \theta = \cos \alpha , θ=(2n+1)α\theta = \left( {2n + 1} \right)\alpha .
For sinθ\sin \theta ,
If sinθ=sinα,θ=nπ+(1)nα\sin \theta = \sin \alpha ,\theta = n\pi + {\left( { - 1} \right)^n}\alpha .
Now, cosθ=cosπ2\cos \theta = \cos \dfrac{\pi }{2}.
Therefore, θ=(2n+1)π2\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}.
Now, sinθ=sin7π6\sin \theta = \sin \dfrac{{7\pi }}{6}.
Therefore, θ=nπ+(1)n7π6\theta = n\pi + {\left( { - 1} \right)^n}\dfrac{{7\pi }}{6}.