Question

Solve sin2 x - sin x - 6 = 0 on the interval [0, 2n).Oπ 11π6 6sin-1 2, sin-13, 2n - sin-1 2, 2n - sin-1 3π 5π3' 3No solution

Answer 1

It is given that sin^2x-sinx-6=0

and the solution is in the interval [0,2n)

Now the equation is quadratic in nature

Therefore solve the equation to get:

sin^2x-sinx-6=0

sin^2x-3sinx+2sinx-6=0

sinx(sinx-3)+2(sinx-3)=0

(sinx+2)(sinx-3)=0

sinx=-2 or sinx=3

Both are not possible since sinx always belongs to [-1,1]

Hence value of -2 or -3 is not valid

Hence the option is D No solution.