Question

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

sin^2x + sin x = 0

Answer 1

what have you tried so far? Usually these types of questions take a little algebra to set up.

Answer 2

i havent been able to do much but sin x = 1/cscx

Answer 3

ok, how about if you factor a sin(x) out of the left side? like this:
\[\sin(x)\cdot(\sin(x)+1) =0\]

Answer 4

ok what about the sin

Answer 5

From algebra class we now know that:
\[\sin(x) = 0 \enspace \text{or} \enspace \sin(x) + 1 = 0\]

Answer 6

sorry, did you have a question about the sin? Which part?

Answer 7

no i figured it out

Answer 8

ok, do you think you can figure the rest out? :)

Answer 9

graph sinx

Answer 10

sure you can do that or if you know the unit circle you can use that too, just find the points where sin is 0 and where sin is -1 and you should be set.

Answer 11

so pi/3 and 5pi/3 right

Answer 12

not, quite, your graph of sin should be like this...

Answer 13

so we are supposed to look in the interval [0, 2pi) which means we should start at x = 0 and go up to 2pi, but don't use 2pi.

Answer 14

so sin(0) = 0 -> x = 0
and sin(pi) = 0 -> x = pi

for the -1 solutions
sin(3pi/2) = -1 -> x = 3pi/2

Answer 15

even though sin(2pi) does equal 0 , we ignore it since it is not in our interval

Answer 16

oh okay thank you

Answer 17

i actually have one more question

Answer 18

Write the expression as the sine, cosine, or tangent of an angle.

cos 96° cos 15° + sin 96° sin 15

Answer 19

it would be cos81