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Question

General solution of x for which cos2x=sinxtanx6 is :

A
nπ±π3,nZ
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B
2nπ±π3,nZ
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C
2nπ±π6,nZ
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D
nπ±π4,nZ
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Solution

The correct option is B 2nπ±π3,nZ
cos2x=sinxtanx66cos3x=1cos2x6cos3x+cos2x1=0
Let cosx=t, where t[1,1]
6t3+t21=0 ...(1)
From the given options, we have
cosπ3=12, cosπ6=32, cosπ4=12

We observe that only t=12 is satisfied the equation (1).
(68+141=0)

Now, after dividing the equation (1) form (2t1), we get
6t3+t21=(2t1)(3t2+2t+1)
So, for remaining roots 3t2+2t+1=0
But Δ<0
No real roots of the quadratic equation.

So, only possible real root of the cubic equation is cosx=12
x=2nπ±π3, nZ

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