Question

The number of roots of the equation cos3x + cos2x = sin3x/2
+ sinx/2
which lie in the interval (0,2π) is​

Answer 1

Answer:

There are 3 roots of the given equations.

Step-by-step explanation:

Given:

$cos\,3x+cos\,2x=sin\,\frac{3x}{2}+sin\,\frac{x}{2}$

To find: Number of Roots of the given equation.

We use the following formula,

$cos\,a+cos\,y=2\:cos\,\frac{a+b}{2}\:cos\,\frac{a-b}{2}$

$sin\,a+sin\,y=2\:sin\,\frac{a+b}{2}\:cos\,\frac{a-b}{2}$

Consider,

$cos\,3x+cos\,2x=sin\,\frac{3x}{2}+sin\,\frac{x}{2}$

$2\:cos\,\frac{3x+2x}{2}\:cos\,\frac{3x-2x}{2}=2\:sin\,{\frac{3x}{2}+\frac{x}{2}}{2}\:cos\,\frac{\frac{3x}{2}-\frac{x}{2}}{2}$

$cos\,\frac{5x}{2}\:cos\,\frac{x}{2}=sin\,\frac{2x}{2}\:cos\,\frac{x}{2}$

$cos\,\frac{5x}{2}\:cos\,\frac{x}{2}=sin\,x\:cos\,\frac{x}{2}$

$cos\,\frac{5x}{2}\:cos\,\frac{x}{2}-sin\,x\:cos\,\frac{x}{2}=0$

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

$\implies cos\,\frac{x}{2}=0\:\:\:and\:\:\:cos\,\frac{5x}{2}-sin\,x=0$

Now Consider,

$cos\,\frac{5x}{2}-sin\,x=0$

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

$-2\:sin\,\frac{{5x}{2}+\frac{\pi}{2}-x}{2}\:sin\,\frac{{5x}{2}-(\frac{\pi}{2}-x)}{2}=0$

$-2\:sin\,\frac{\frac{\pi}{2}+\frac{3x}{2}}{2}\:sin\,\frac{\frac{5x}{2}-\frac{\pi}{2}+x}{2}=0$

$-2\:sin\,\frac{\frac{3x}{2}}{2}\:sin\,\frac{-(\frac{\pi}{2}-\frac{7x}{2})}{2}=0$

$2\:sin\,\frac{3x}{4}\:sin\,\frac{\frac{\pi}{2}-\frac{7x}{2}}{2}=0$

$2\:sin\,\frac{3x}{4}\:sin\,\frac{\frac{7x}{2}}{2}=0$

$2\:sin\,\frac{3x}{4}\:sin\,\frac{7x}{4}=0$

$\implies sin\,\frac{3x}{4}=0\:\:\:and\:\:\:sin\,\frac{7x}{4}=0$

Therefore, There are 3 roots of the given equations.

Answer 2

Step-by-step explanation:

hope it helps you......