For `x in (0,pi),` the equation `sinx+2`sin` 2x-sin3x=3` has (A)infinitely many solutions (B)three solutions (C)one solution (D)no solution

Options

A. infinitely many solution
B. three solutions
C. one solutions
D. no solution

Answer

d

Explanation

Plant For solving this type of questions, obtain the LHS and RHS in equation and examine, the two are equal or not for a given interval.
Given, trigonometrical equation
` (sinx - sin 3x) + 2 sin 2x = 3`
` rArr - 2 cos 2 x sin x + 4 sin x cos x = 3 `
` " " [ because sin C - sin D = 2 cos ((C +D)/(2)) sin ((C-D)/(2)) and sin 2 theta = 2sin theta cos theta ]`
`rArr 2 sinx (2 cos x - cos 2 x ) = 3`.
` rArr 2 sin x ( 2 cos x - 2 cos^(2) x+ 1)=3`
`rArr 2 sin x [(3)/(2)- 2 (cos x - (1)/(2))^(2)] = 3`
` rArr" " 3 sin x - 3 = 4 (cos x - (1)/(2)) ^(2) sinx `
As ` x in (0, pi ) " " ` LHS ` le ` 0 and RHS ` ge ` 0
For solution to exist, LHS = RHS =0
Now, LHS = 0
` rArr " " 3 sinx - 3 = 0 `
` rArr" " sin x = 1 `
` rArr " " x = (pi)/(2)`
For `" " x = (pi)/(2)`
RHS = ` 4 (cos ""(pi)/(2) - (1)/(2))^(2) sin ""(pi)/(2) = 4 (1)/(4)(1) = 1 ne 0 `
` therefore ` No solution of the equation exists.