NTA Abhyas JEE Main2020MathematicsTrigonometric EquationsPractice
The general solution of the system of equations sin 3 x + sin 3 2 π 3 + x + sin 3 4 π 3 + x + 3 4 cos 2 x = 0 and cos x ≠ 0 is
Options
- Ax = 2 k + 1 π 10 , k ∈ Z
- Bx = 2 k + 1 π 5 , k ∈ Z
- Cx = 4 k + 1 π 10 , k ∈ Z
- Dx = 4 k + 1 5 π , k ∈ Z
Correct answer
C. x = 4 k + 1 π 10 , k ∈ Z
Step-by-step solution
∵ sin 3 x = 3 sinx - 4   sin 3 x    ⇒ sin 3 x = 1 4 3 sinx - sin 3 x     The given equation reduces to 1 4 3 sinx - sin 3 x + 1 4 3 sin 2 π 3 + x -   sin 2 π + 3 x + 1 4 ( 3 sin 4 π 3 + x - sin 4 π + 3 x + 3 4 cos 2 x = 0   3 sin ⁡ x + sin 2 π 3 + x + sin ⁡ 4 π 3 + x - sin ⁡ 3 x + sin ⁡ 2 π + 3 x + sin ⁡ 4 π + 3 x + 3 cos ⁡ 2 x = 0 3   sinx + 2 sinx   cos ⁡   2 π 3 - 3