Trigonometry> Find the value of x between 0 and 2 pi sa...

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

Shefali , 7 Years ago

Grade 11

3 Answers

Deepak Kumar Shringi

Last Activity: 7 Years ago

Aarushi Ahlawat

Last Activity: 7 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

$cos(3x)+cos(2x)=sin\left ( \frac{3x}{2} \right )+sin\left ( \frac{x}{2} \right )$

$2cos\left ( \frac{3x+2x}{2} \right )cos\left ( \frac{3x-2x}{2} \right )=2sin\left ( \frac{\frac{3x}{2}+\frac{x}{2}}{2} \right )cos\left ( \frac{\frac{3x}{2}-\frac{x}{2}}{2} \right )$

$cos\left ( \frac{5x}{2} \right )cos\left ( \frac{x}{2} \right )=sin\left ( x \right )cos\left ( \frac{x}{2} \right )$

$cos\left ( \frac{x}{2} \right ) \left ( cos\left ( \frac{5x}{2} \right )- sin\left ( x \right ) \right )=0$

$cos\left ( \frac{x}{2} \right ) \left ( cos\left ( \frac{5x}{2} \right )- cos\left (\frac{\pi }{2} - x \right ) \right )=0$

$cos\left ( \frac{x}{2} \right ) \left ( cos\left ( \frac{5x}{2} \right )- cos\left (\frac{3 \pi }{2} + x \right ) \right )=0$

Now the solutions:

$cos\left ( \frac{x}{2} \right ) = 0$

$x=\pi$

$cos\left ( \frac{5x}{2} \right )-cos\left ( \frac{\pi}{2}-x \right )=0$

$x=\frac{\pi}{7}$

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

Last Activity: 7 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.