Question

The no. Of real equation 2sin3x+sin7x-3=0 which lie in the interval (-2pi,2pi) is

Answer 1

By looking at the expression,

⟹sin(3x)=1and

sin(7x)=1

⟹x∈(4n+1)π6∩(4m+1)π14

Where, n∈Z

And m∈Z

So n and m can have following values for the given range.

m=0,±1,±2,±3,±4,±5,±6,−7⟹x∈(π/14,5π/14,9π/14,13π/14,17π/14,3π/2,25π/14,−π/2,−11π/14,−15π/14,−19π/14,−23π/14,−27π/14)

n=0,±1,±2,−3⟹x∈(π/6,−π/2,5π/6,3π/2,−7π/6,−11π/6)

It can be seen that only (3π/2,−π/2)lies inside the region of choice .

Answer 2

The given equation is , 2 sin 3 x + sin 7 x -3=0

This is possible when , sin 3 x=1 and sin 7 x =1 as→ -1≤ sin x ≤ 1.

→Sin 3 x=1

→Sin 3 x = Sin π/2

→3 x =n π + $(-1)^{n}\frac{\pi }{2}$

→ x = $\frac{n  \pi + (-1)^{n}\frac{\pi }{2}}{3}$ where n is an integer.

For, n=-6,-5,-4,-3,-2,-1,0,1,2,3,4,5 we get x= -11π/6, -11π/6,-7π/6,-7π/6,

-π/2, -π/2,π/6, π/6, 5π/6, 5π/6,3π/2, 3π/2,

Similarly , sin 7 x =1

→ Sin 7 x = Sin π/2

→x = $\frac{n \pi + (-1)^{n}\frac{\pi }{2}}{7}$ where n is an integer.

            →    The values of x are    π/14,π/14,5π/14,5π/14,9π/14,9π/14,13π/14,13π/14,17π/14,17π/14,3π/2,3π/2,25π/14, 25π/14,27π/14, 27π/14. for x=0,1,2,3,4,5,6,7,8,9,10,11,12,13,14 .

Also for x=-1,-2,-3,-4,-5,-6,-7,-8,-9,-10,-11,-12,-13,-14

The solution are , -3π/14,-3π/14, -π/2,-π/2,-11π/14,-11π/14,-15π/14,-15π/14,-19π/14,-19π/14,-23π/14,-23π/14,-27π/14,27π/14.

So, the common values of x are -π/2,and  3π/2.

So,there are two solution[ -π/2,and  3π/2] of real equation →2 sin3x+ sin7x-3=0 which lie in the interval (-2π,2π).