Question

13. Solve cos 3x + cos2x = sin 3x/2 + sin x/2 ,0< x < 2.​

Answer 1

Answer:

I hope it's helps you

Step-by-step explanation:

cos3x+cos2x=sin(3x/2)+sin(x/2)

∴2cos

2

5x

cos

2

x

−2sinxcos

2

x

=0

∴2cos(x/2)[cos(5x/2)−sinx]=0

cos(

2

x

)=0 ∴

2

x

=(n+

2

1

)π

or x=2nπ+π,

n=0, x=π as 0≤x≤2π

cos

2

5x

=sinx=cos(

2

π

−x)

2

5x

=2nπ±(

2

π

−x)

Taking +ive sign (5x/2)+x=2nπ+π/2,

(7x/2)=(4n+1)(π/2)

∴x=(4n+1)π/7

n=0,x=π/7,n=1,x=5π/7,n=2,x=9π/7,

n=3,x=13π/7

Now taking −ive sign,

(5x/2)−x=2nπ−π/2

or (3x/2)=(4n−1)π/2

∴x=(4n−1)π/3,

∴ For n=1, x=π.

∴x=

7

π

,

7

5π

,π,

7

9π

,

7

13π

.