Q no. 27. √3cosx - sinx = cos^10y + sec^10y.Find the general solution.
Q no. 27. √3cosx - sinx = cos^10y + sec^10y.
Find the general solution.
Aniket Chatterjee{ANI} , 7 Years ago
Grade 12th pass
Trigonometry> Q no. 27. √3cosx - sinx = cos^10y + sec^1...
1 Answers
Aditya Gupta
Let cos^10y= z which obviously implies z>0
Then RHS= z+1/z >=2
But LHS
So for them to be equal, both should be 2.
So z+1/z= 2 or z=1 or cosy= +-1 or y = mπ
And sin(pi/3-x)=1= sin(π/2)
So x= 2nπ-π/6
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