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solve for x sin3x *sin(cube) x +cos3x *cos(cube) x=0 arun sir answered me but it was a tough method i need some easy way as i am in 11 th so a method which i couldget please answer>>>

solve for x
sin3x *sin(cube) x +cos3x *cos(cube) x=0
arun sir answered me but it was a tough method i need some easy way as i am in 11th so a method which i couldget 
please answer>>>

Grade:11

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
sin3x.sin^{3}x + cos3x.cos^{3}x = 0
sin3x.sin^{3}x = -cos3x.cos^{3}x
tan3x = \frac{-1}{tan^{3}x}
\frac{3tanx-tan^{3}x}{1-3tan^{2}x} = \frac{-1}{tan^{3}x}
3tan^{4}x-tan^{6}x = -1+3tan^{2}x
tan^{6}x -3tan^{4}x + 3tan^{2}x -1 = 0
It is a simple polynomial, you can easily solve. And if you have options, you can hit & trial.
tan^{6}x - 1 = 3tan^{4}x - 3tan^{2}x
(tan^{3}x - 1)(tan^{3}x + 1) = 3tan^{2}x(tan^{2}x-1)
tanx = \pm 1
x = n\pi \pm \frac{\pi }{4}
rashmi
22 Points
7 years ago
sir can u explain me how u solved the third last step not getting ?? is there any such identity??

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