Question

Help please ? =)

Answer 1

well part I just says to replace "sin(x)" with "u" so lmk what you get when you do that

Answer 2

so just sin (u)?

Answer 3

not quite, "u" replaces the entire sin(x) expression

so sin^2(x) becomes u^2

Answer 4

2sin^2 x-5 u-3=0

Answer 5

almost, 2u^2 - 5u - 3 = 0 and that's it for part I

Answer 6

then for part II just factor 2u^2 - 5u - 3 = 0 like you normally would

Answer 7

u=3,-1/2

Answer 8

good but I think part II only wants the factored form

(u-3)(2u+1) = 0 as the sol'n

then for part III go ahead and write down your two u values

Answer 9

then for part IV replace "u = 3" and "u = -1/2" with "sin(x) = 3 and sin(x) = -1/2" since they want you to replace u with sin(x)

Answer 10

so sin(x) = 3 and sin(x) = -1/2 for part IV

Answer 11

yes

Answer 12

then for part V solve for x using the unit circle to see where sin(x) = -1/2

sin(x) = 3 does not have a solution since 3 is outside the range of sin

Answer 13

so part V is just sin(x) = 3 does not have a solution since 3 is outside the range of sin

Answer 14

you also need to solve for sin(x) = -1/2

Answer 15

7pi/6

Answer 16

good but there's one more

Answer 17

11pi/6

Answer 18

wait so these two answers for part V

Answer 19

good so your sol'ns for part V are 7pi/6 +/- 2pi *n and 11pi/6 +/- 2*pin

Answer 20

it wouldn't hurt to also include the part about sin(x) = 3 not having a sol'n

Answer 21

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
then for part V solve for x using the unit circle to see where sin(x) = -1/2

sin(x) = 3 does not have a solution since 3 is outside the range of sin
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 22

yeah that

Answer 23

yeah thats what I was going to ask would i include that statement in part V or IV

Answer 24

I'll go with part V ^^

Answer 25

yeah it's part V