Question

solve the equation in the interval [0,2pi)

sinx-2sin^2x=0

Answer 1

Factorize using a common factor of sin x then.....

Answer 2

ok i got sinx(1-2sinx)=0

Answer 3

Excellent, you now have two things that multiply to give a result of zero .........

Answer 4

sinx=0 sinx= 1/2

Answer 5

top result . Do you use "exact values"?

Answer 6

yes and thats what i really have trouble with

Answer 7

Do you use a little triangle to help you remember exact values?

Answer 8

\[\sin x-2\sin ^{2}x=0\]
factor sinx
\[\sin x \left( 1-2\sin x \right)=0\]
equate each factor to zero
(a) \[\sin x=0\] or (b)\[1-2\sin x=0\]
the solution to (a) is \[x=0,\pi\]
the solution to (b) is
\[x=\pi/6,5\pi/6\]

Answer 9

like with the unit circle?

Answer 10

like this (but neater)

Answer 11

ok i understand how do you get the second angle the \[\frac{ 5\pi }{ 6}\]

Answer 12

Thats where the unit circle comes in. Sin is positive in two quadrants.
more pictures help?

Answer 13

yes please :)

Answer 14

I'll try