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Question

Solve 2cos2θ+cosθ1=0


A

2nπ nN

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B

2nππ3 nN

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C

2nππ6 nN

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D

(2n+1)π6 nN

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Solution

The correct option is B

2nππ3 nN


We can see that the given expression is a quadratic in cos. We can find its roots and find the general solutions easily.

cosθ=11+84
cosθ=12 or cosθ=1

We know that the general solution forumla for the equation
cosx=cosα
is x=2nπ±α
θ=2nπ±π3 or π=(2n±1)π


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