184k views
3 votes
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

2 Answers

6 votes

Answer:


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

User Lfagundes
by
7.5k points
4 votes

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

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.

User Ulfhetnar
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories