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`        find the general solution of the equation 7cos2x+3sin2x=4`
6 months ago

Arun
22342 Points
```							7 cos²x + 3 sin²x = 4 cos²x + 3 cos²x + 3 sin²x = 4 cos²x + 3 (cos²x + sin²x) = 4 cos²x + 3 This can only be = 4 when 4 cos²x = 1, i.e. for only certain values of x. However, an identity (which you want us to prove) should be true for all values of x. So I believe that instead of this being an identity that needs to be proven, this is actually an equation that needs to be solved. Please be careful how you phrase your questions. 7 cos²x + 3 sin²x = 4 4 cos²x + 3 cos²x + 3 sin²x = 4 4 cos²x + 3 (cos²x + sin²x) = 4 4 cos²x + 3 = 4 4 cos²x = 1 cos²x = 1/4 cosx = ±1/2 x = π/3 + πk, 2π/3 + πk
```
6 months ago
prince sunjot dutt
58 Points
```							7 cos²x+3sin²x=47cos²x+3(1-cos²x)=47cos²x+3-3cos²x=44cos²x=1Cosx=±1/2Wkt:ifCosx=cosaThen, x=2nπ±a,______n€ICosx=cos(π/3) or Cosx=-cos(π/3)X=2nπ±π/3       or x= 2nπ± 2π/3                    n={±1,±2,±3,±4,±5.........)
```
6 months ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions