Let a, b, c, d ∈ R. Then the cubic equation of the type ax 3 + bx 2 + cx + d = 0 has either one root real or all three roots are real. But in case of trigonometric equations of the type a sin 3 x + b sin 2 x + c sinx + d = 0 can possess several solutions depending upon the domain of x.To solve an equation of the type a cos θ + b sin θ = c. The equation can be written as
cos ( θ – α ) = c/ 
.
The solution is θ = 2n π + α ± β , where tan α = b/a, cos β = c/ 
.
(i)On the domain [– π , π ] the equation 4sin 3 x + 2 sin 2 x – 2sinx – 1 = 0 possess
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) 4sin 3 x + 2 sin 2 x – 2sinx – 1 = (2sin x + 1) (2sin 2 x – 1) = 0
∴ sinx = – 
, ± 
∴ there are 6 solutions.
(ii) 3 = cos 4x + 
= cos 4x + 5 sin 2x
i.e 3 = 1 – 2 sin 2 2x + 5 sin 2x
i.e sin 2x = 
∴ 2x = 
, 
Thus there are two solutions.
(iii) (i) when tan x ≥ 0, then the equation becomes tan x = tan x + 
i.e 
= 0 (not possible)
(ii) when tan x < 0, then the equation becomes – tan x = tan x + 
i.e sin x = – 
∴ x = 
is the only solution.
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems