Q no. 27. √3cosx - sinx = cos^10y + sec^10y. Find the general solution.

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Aditya Gupta · Answer

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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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.

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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.

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Q no. 27. √3cosx - sinx = cos^10y + sec^10y.
Find the general solution.