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Question: Solve \[\sin 2\theta + {\text{cos 2}}\theta \] = 0 This question has multiple correct options A....

Solve sin⁡2θ+cos 2θ\sin 2\theta + {\text{cos 2}}\theta = 0
This question has multiple correct options
A.θ=2nπ−π3,n∈Z\theta = 2n\pi - \dfrac{\pi }{3},n \in Z
B.θ=2nπ3−π4,n∈Z\theta = \dfrac{{2n\pi }}{3} - \dfrac{\pi }{4},n \in Z
C.θ=2nπ−π2,n∈Z\theta = 2n\pi - \dfrac{\pi }{2},n \in Z
D.θ=2nπ3−π6,n∈Z\theta = \dfrac{{2n\pi }}{3} - \dfrac{\pi }{6},n \in Z

Explanation

Solution

Before attempting this question you should have prior knowledge about the trigonometric equations and also remember to usecos⁡(π2±θ)=∓sin⁡θ\cos \left( {\dfrac{\pi }{2} \pm \theta } \right) = \mp \sin \theta , use this information to approach the solution of the question.

Complete step by step solution:
According to the given information we have trigonometric equation sin⁡2θ+cos 2θ\sin 2\theta + {\text{cos 2}}\theta = 0
Simplifying the above equation we get
cos⁡2θ=−sin⁡2θ\cos 2\theta = - \sin 2\theta
Since we know that cos⁡(π2+θ)=−sin⁡θ\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta
Therefore cos⁡2θ=cos⁡(π2+θ)\cos 2\theta = \cos \left( {\dfrac{\pi }{2} + \theta } \right) (equation 1)
Since we know that trigonometric equation i.e. cos⁡θ=cos⁡α\cos \theta = \cos \alpha and the general solution is given as θ=2nπ±α,n∈Z\theta = 2n\pi \pm \alpha ,n \in Z
Using the above trigonometric equation in equation 1 we get
θ=2nπ±(π2+2θ),n∈Z\theta = 2n\pi \pm \left( {\dfrac{\pi }{2} + 2\theta } \right),n \in Z
Now considering the positive sign for above equation we get
θ=2nπ+(π2+2θ),n∈Z\theta = 2n\pi + \left( {\dfrac{\pi }{2} + 2\theta } \right),n \in Z
⇒\Rightarrow θ−2θ=2nπ+π2,n∈Z\theta - 2\theta = 2n\pi + \dfrac{\pi }{2},n \in Z
⇒\Rightarrow θ=−(2nπ+π2),n∈Z\theta = - \left( {2n\pi + \dfrac{\pi }{2}} \right),n \in Z
Now taking negative sign for equation θ=2nπ±(π2+2θ),n∈Z\theta = 2n\pi \pm \left( {\dfrac{\pi }{2} + 2\theta } \right),n \in Z
We get θ=2nπ−(π2+2θ),n∈Z\theta = 2n\pi - \left( {\dfrac{\pi }{2} + 2\theta } \right),n \in Z
θ+2θ=2nπ−π2,n∈Z\theta + 2\theta = 2n\pi - \dfrac{\pi }{2},n \in Z
⇒\Rightarrow 3θ=2nπ−π2,n∈Z3\theta = 2n\pi - \dfrac{\pi }{2},n \in Z
⇒\Rightarrow θ=2nπ3−π6,n∈Z\theta = \dfrac{{2n\pi }}{3} - \dfrac{\pi }{6},n \in Z
Hence option D is the correct option.

Note: The above question was totally based on the concept of trigonometry and its identities which can be explained as the concept which relates the sides and the angle of a right angled triangle whereas the trigonometric identities are equalities which consists of trigonometric identities like sin theta, cos theta, etc. which are true for every occurring variables in such cases where sides of both are well defined.