Question

find all solutions to the equation:
(sin x)(cos x)=0

Answer 1

@SolomonZelman ,do u know how 2 solve this one?

Answer 2

Warning: \(\normalsize\color{red}{ \rm{It~~~will~~~be~~~awkward~!} }\)
sin(x) cos(x)=0
[ sin(x) cos(x) ]²=0²
sin²x cos²x =0
sin²x ( 1 - sin²x) = 0
sin\(\small\color{black}{ ~^{4} }\)x - sin²x =0
sin\(\small\color{black}{ ~^{4} }\)x = sin²x

can you finish ?

Answer 3

multiplication of two numbers gives zero then both of these no.are itself zero i.e

sinX=0 and cosX=0

now solve both of these separately,,,

Answer 4

sin²x = 1 (when you divide both sides by `sin²x`)

Answer 5

sin(x)=1 or -1

sin\(\small\color{black}{ ~^{-1} }\)1 or sin\(\small\color{black}{ ~^{1} }\)1

Answer 6

No I don't know how to solve this :P

Answer 7

its ok @SolomonZelman .thanks alot:)

Answer 8

the way I did it was really retarded, BUT algebraic.

Answer 9

sinX=0 and cosX=0
x=0,pi and x=pi/2,3pi/2
but,how can i get the general solutions?@zaibali.qasmi

Answer 10

lol=p

Answer 11

@zaibali.qasmi ?

Answer 12

so these are four solutions regarding this eq...
for general sol just sum up these it will be from 0 to pi/..

if it is zero then eq satisfy and if it is pi then again it is satisfying the result..

Answer 13

thankyou,but I also need the general solutions

Answer 14

it should be [0,pi]

Answer 15

this is one of the options:
pi/2+npi,npi
so.is this the correct answer?

Answer 16

what are the option ...??all..

Answer 17

a)npi
b)pi/2+npi,npi
c)pi/2+2npi
d)pi/2+npi