Question

help please medal and fan
Solve the equation on the[0,2π).
interval ​[0,2piπ​).

18sinx=8cos^2 x−15

Answer 1

hint : if you write cos^2 x in terms of sin^2 x,
you will get a quadratic equation in sin x :)

Answer 2

I dont understand

Answer 3

how will you write cos^2 x in terms of sin^2 x?
any pythogorean identity?

Answer 4

i am completely lost .. i really need that step to step process

Answer 5

did you know about \(\sin^2 x+\cos^2 x =1\)
?

Answer 6

Yes

Answer 7

1-sin2x=cos2x

Answer 8

what is cos^2 x from that ?

Answer 9

theres that one too i think

Answer 10

you mean
\(\cos^2 x = 1- \sin^2 x\)
?
yes, you need to use this and replace cos^2 x by the right side

Answer 11

yes thats what i meant :)

Answer 12

18sinx=8(1-sin^2 x)−15

let u = sin x

Answer 13

18u=8(1-u^2)-15

Answer 14

this is really stressing

Answer 15

i dont understand

Answer 16

but did you get this:
18sinx=8(1-sin^2 x)−15
?

Answer 17

yes u changed it it ... and it equals the same

Answer 18

now i just made a substitution, for simplicity

writing sin x as u

Answer 19

18sinx=8(1-sin^2 x)−15

18u=8(1-u^2)−15

this is a quadratic in u

Answer 20

Okay

Answer 21

can you solve quadratics?

Answer 22

no

Answer 23

have a look!
http://www.regentsprep.org/regents/math/algtrig/ate3/quadlesson.htm

Answer 24

okay