Question

Find all solutions to the equation.

Answer 1

Answer:

$x=n\pi$ or $x= \frac{(2n\pm1)\pi}{2}$, where $n\ge0$

Step-by-step explanation:

The given trigonometric equation is:

$(\sin x)(\cos x)=0$

By the zero product principle;

Either $\sin x=0$ or $\cos x=0$

When  $\sin x=0$,  then $x=n\pi$

When $\cos x=0$, then $x=2n\pi \pm \cos^{-1}(0)$

This implies that: $x=2n\pi \pm \frac{\pi}{2}$

Therefore the general solution is  $x=n\pi$ or $x= \frac{(2n\pm1)\pi}{2}$, where $n\ge0$

Answer 2

Answer:

B. pi divided by two plus n pi comma n pi such that n equals zero, plus or minus one, plus or minus two to infinity

Step-by-step explanation:

Given equation,

$(sin x)(cosx)=0$

$\implies sinx=0\text{ or }cosx=0$

$x=sin^{-1}(0)\text{ or }x=cos^{-1}(0)]$

$x=\pi, 2\pi, 3\pi,......\text{ or }x=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}....$

$\implies x = n\pi\text{ or }x=\frac{\pi+2n\pi}{2}$

Or

$x=n\pi\text{ or }x= \frac{\pi}{2}+n\pi$

Where, n = 0, 1, -1, -2, ........∞

Hence, option 'B' is correct.