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```				   solve the equation

cos3X sin3(cube)X + sin3X cos3(cube)X  =  0
```

6 years ago

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```										Dear somesh,
we know that,
cos^3x=(cos3x+3cosx)/4 and  sin^3x=(3sinx-sin3x)/4                    [since cos3x=4cos^x-3cosx]
substuting in the given equation
cos3X sin3(cube)X + sin3X cos3(cube)X  =  0
cos3x{(3sinx-sin3x)/4}+sin3x{(cos3x+3cosx)/4} = 0
{3sinxcos3x-cos3xsin3x+sin3xcos3x+3cosxsin3x}/4 = 0
3/4(sinxcos3x+cosxsin3x) = 0
sin4x = 0
Hence x=n∏/4
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

```
6 years ago

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