Question

find all solutions.
7 sin2x - 14 sin x + 2 = -5

Answer 1

rearrange the equation to equal 0
and let sin(x) = s
7s^2 - 14s +7 =0
you can divide through by 7
s^2-2s+1=0

then factorize
s=-1 (repeated)
sin(x)=-1
solve for x

Answer 2

I dont get it, because when i make it = to zero i get
7sin^2x-9sinx+7=0 then i cant divide by 7 through it..

Answer 3

okay, the -5 only affect the '2' because the '-14sinx' has the sin(x) attached to it

Answer 4

in the same way you don't alter the sin^2(x) bit

Answer 5

so would i just subtract 2 then divide by 7? after that i would get sin^2 x-14sinx=-1
then what

Answer 6

you also need to divide the 14sin(x)
so you would have
sin^2(x)-2sin(x)=-1
so sin^2(x)-2sin(x)+1=0

Answer 7

oh okay that makes sense. then whats the next step to find the solutions?

Answer 8

i just need to know what the next steps are to finish the problem

Answer 9

@Haseeb96 can you help?

Answer 10

I told you at the top of this thread! ^^

Answer 11

You just solve sin(x)=-1
therefore
x=sin^-1 (-1) = x

Answer 12

since, sin2x=2sinxcosx
7(2sinxcosx)-14sinx+2=-5
14sinxcosx-14sinx+2=-5
14sinx(cosx-1)=-5-2
sinx(cosx-1)=-7\14
sinx(cosx-1)=-1\2
either ,
sinx=-1\2 or cosx-1=-1\2
sinx=-1\2 or cosx=-1\2 +1
sinx =-1\2 or cosx=1\2
\[x=\sin^{-1} \frac{ -1 }{ 2 } or x=\cos^{-1} \frac{ 1 }{ 2 }\]

since, sin 210 = -1\2 and cos 60=1\2
x=240 or x= 60
\[x=\frac{ 4\pi }{ 6 } or x=\frac{ \pi }{ 3 }\]

Answer 13

thank you so much !

Answer 14

oh I'm sorry, I thought it was sin(^2)x not sin(2x) I apologise!
The problem with denoting things differently. My bad