Question

Let $\theta ϵ[0,4\pi ]$ satisfy the equation $(\sin \theta +2)(\sin \theta +3)(\sin \theta +4)=6$. If the sum of all the values of $\theta$ is of the form $k\pi$ then the value of $k$ is

Answer 1

$\Rightarrow (\sin \theta +2)(\sin \theta +3)(\sin \theta +4)=6\Rightarrow (\sin {}^2\theta +5\sin \theta +6)(\sin \theta +4)=6\Rightarrow \sin {}^3\theta +4\sin {}^2\theta +5\sin {}^2\theta +20\sin \theta +6\sin \theta +24=6\Rightarrow \sin {}^3\theta +9\sin {}^2\theta +26\sin \theta +18=0let,\sin \theta =x{x}^3+9x^2+26x+18=0(x+1)(x^2+8x+18)=0(x+1){(x+3\sqrt{2})}^2=0(\sin \theta +1)=0,(\sin \theta +3\sqrt{2})=0\sin \theta =-1,\sin \theta =-3\sqrt{2}\therefore \theta =\frac{3\pi }{2}\pm 2\pi n=\pi (\frac{3}{2}\pm 2n)\therefore k=\frac{3}{2}\pm 2n$