Question
Question: How do you solve \(\sin 2\theta + \cos \theta = 0\) where, \(0 < \theta < 2\pi \)?...
How do you solve sin2θ+cosθ=0 where, 0<θ<2π?
Solution
In this question, we are given a trigonometric equation and the limit between which θ 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 0. Then, find the value or values of θ. Use the given limit to find the required values.
Formula used: sin2θ=2sinθcosθ
Complete step-by-step solution:
The trigonometric equation given to us is sin2θ+cosθ=0.
Firstly, expand the first term of the given equation using a basic trigonometric identity.
⇒sin2θ+cosθ=0
Using sin2θ=2sinθcosθ,
⇒2sinθ.cosθ+cosθ=0
Taking cosθ common,
⇒cosθ(2sinθ+1)=0
Now, we have two factors. We will keep each factor equal to 0.
⇒cosθ=0,2sinθ+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 is equal to 0 at 2π and 23π(=π+2π).
Hence, θ=2π,23π.
Now, we will solve the second part.
⇒2sinθ+1=0
⇒sinθ=2−1
We know that sinθ is negative in the third and fourth quadrant. Therefore, our required values will also lie in those quadrants only. We know that sinθ is equal to 21 at 6π .
Therefore, its negative will be on 67π(=π+6π) and 611π(=2π−6π).
Hence, θ=2π,23π,67π,611π.
Note: What if the range within with θ 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=2−1
We can write it as –
⇒cosθ=cos2π,sinθ=sin67π
Putting them in their general formula –
For cosθ,
If cosθ=cosα, θ=(2n+1)α.
For sinθ,
If sinθ=sinα,θ=nπ+(−1)nα.
Now, cosθ=cos2π.
Therefore, θ=(2n+1)2π.
Now, sinθ=sin67π.
Therefore, θ=nπ+(−1)n67π.