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Find the value of x between 0 and 2 pi satisfying Cos3x + cos2x = sin 3x/2 + sin x/2

Find the value of x between 0 and 2 pi satisfying 
Cos3x + cos2x = sin 3x/2 + sin x/2

Grade:11

3 Answers

Deepak Kumar Shringi
askIITians Faculty 4405 Points
3 years ago
562-825_Capture.PNG
Aarushi Ahlawat
41 Points
3 years ago
First convert both the sides in multiplication form and then take common to get two factors. Now find the solutions for each of the factor over the given domain
.
cos(3x)+cos(2x)=sin\left ( \frac{3x}{2} \right )+sin\left ( \frac{x}{2} \right )
2cos\left ( \frac{3x+2x}{2} \right )cos\left ( \frac{3x-2x}{2} \right )=2sin\left ( \frac{\frac{3x}{2}+\frac{x}{2}}{2} \right )cos\left ( \frac{\frac{3x}{2}-\frac{x}{2}}{2} \right )
 
cos\left ( \frac{5x}{2} \right )cos\left ( \frac{x}{2} \right )=sin\left ( x \right )cos\left ( \frac{x}{2} \right )
cos\left ( \frac{x}{2} \right ) \left ( cos\left ( \frac{5x}{2} \right )- sin\left ( x \right ) \right )=0
cos\left ( \frac{x}{2} \right ) \left ( cos\left ( \frac{5x}{2} \right )- cos\left (\frac{\pi }{2} - x \right ) \right )=0
OR
cos\left ( \frac{x}{2} \right ) \left ( cos\left ( \frac{5x}{2} \right )- cos\left (\frac{3 \pi }{2} + x \right ) \right )=0
Now the solutions:
If
 
cos\left ( \frac{x}{2} \right ) = 0
x=\pi
 
if
cos\left ( \frac{5x}{2} \right )-cos\left ( \frac{\pi}{2}-x \right )=0
x=\frac{\pi}{7}
and if
 
cos\left ( \frac{5x}{2} \right )-cos\left ( \frac{3\pi}{2}+x \right )=0
 
x=\pi
 
Hence possible values of x between 0 and 2pi are   
 
\frac{\pi}{7} ~ and ~ \pi
 
 
 
Aarushi Ahlawat
41 Points
3 years ago
Above mentioned general solution in answer 1 are not correct for all values of n and for + and -.
 
All the solution between 0 and 2pi are pi/7, 5pi/7, pi, 9pi/7 and 13pi/7 only. 
 
One must consider the validity conversion of sin(x) term to cos in last step. It must remain valid for both the cases instead of only for cos cases. Hence half of general solution for cos will not be correct solutions. 
 
Validate yourself for so obtained solutions.

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