Question

Find all solutions to the equation.

(sin x)(cos x) = 0


A. n pi such that n equals zero, plus or minus one, plus or minus two to infinity


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


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


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

Answer 1

`x=n\pi` or
`x= ((2n\pm1)\pi)/(2)`, where
`n\ge0`

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 (\pi)/(2)`

Therefore the general solution is
`x=n\pi` or
`x= ((2n\pm1)\pi)/(2)`, where
`n\ge0`

Answer 2

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

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=(\pi)/(2), (3\pi)/(2), (5\pi)/(2)....`

`\implies x = n\pi\text{ or }x=(\pi+2n\pi)/(2)`

Or

`x=n\pi\text{ or }x= (\pi)/(2)+n\pi`

Where, n = 0, 1, -1, -2, ........∞

Hence, option 'B' is correct.