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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.

```
2 years ago

2074 Points
```							Let cos^10y= z which obviously implies z>0Then RHS= z+1/z >=2But LHS So for them to be equal, both should be 2.So z+1/z= 2 or z=1 or cosy= +-1 or y = mπAnd sin(pi/3-x)=1= sin(π/2)So x= 2nπ-π/6
```
2 years ago
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• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions