Question

Find all solutions in the interval [0, 2pi).

(sin x) (cos x) = 0

Answer 1

set sinx = 0 or cosx=0
this is a start, do u know what angles to get sinx=0, cosx=0 ?

Answer 2

sin =0 and cos=pi/2

Answer 3

sinx=0, it satisfied for x=0, x=pi, x=2pi
now, find the solution for cosx=0 ?

Answer 4

pi/2, 3pi/2?

Answer 5

opss.. nevermind for x=2pi, it didnt including in interval [0,2pi) :)
yes, u are right

Answer 6

Thank you! do you think you can help me with my last one?
its kind of tricky?

Answer 7

post it

Answer 8

on here? or close this one and make another?

Answer 9

just here

Answer 10

Find all solutions to the equation.

7 sin2x - 14 sin x + 2 = -5

Answer 11

for this time u must fator out, first...
7 sin2x - 14 sin x + 2 = -5 or
7 sin2x - 14 sin x + 2 + 5 = 0
7 sin2x - 14 sin x + 7= 0
simplify, divided by 7 gives
sin^2 x - 2sinx + 1 = 0
now, can u factor out this

Answer 12

and what happens to the 7?

Answer 13

if the equation of 7 sin^2 x - 14 sin x + 7= 0 divided by 7 to both sides, it means
7/7 sin^2 x - 14/7 sin x + 7/7= 0/7 u get
sin^2 x - 2sin x + 1= 0 right ?

Answer 14

ok you can't just factor it out?

Answer 15

i can, can u ? :)

Answer 16

yea....? :/ but if you factor it out, what happens to the 7?

Answer 17

so x= pi/2, 3pi/2?

Answer 18

u dont need the 7 again, it passed away :)

Answer 19

ok is my top answer correct? :)

Answer 20

finally, the answer will be x=pi/2 only
do u know why ?

Answer 21

no....?

Answer 22

why not 3pi/2?

Answer 23

look this:
from sin^2 x - 2sin x + 1= 0, factor out gives
(sinx - 1)(sinx - 1)=0
it satisfied for sinx - 1 = 0, solve for x

Answer 24

sinx=1
so it equals pi/2
oh because it is not asking for a negative? and 3pi/2=-1?

Answer 25

yeah, u are right :)

Answer 26

oh ok! thank you!!

Answer 27

very welcome :)