Question

Find all solutions to the equation.

7 sin2x - 14 sin x + 2 = -5

Answer 1

Not sure if you want to solve:
\(7\sin^2(x)-14\sin(x)+2=-5\)
or
\(7\sin(2x)-14\sin(x)+2=-5\)
In either case, factor out the common factor 7 to simplify things a little.

Answer 2

sorry the first one

Answer 3

But I can't factor out 7 because of the 2

Answer 4

first put it in standard form by adding +5 to both sides
\[ 7\sin^2(x)-14\sin(x)+2=-5 \\ 7\sin^2(x)-14\sin(x)+7=0\]

Answer 5

You can always let sinx = x then factor normally and plug back sinx to find the solution

Answer 6

\[7x^2-14x+2=5\]

Answer 7

I'm sorry. Are you saying all the possible solutions are x=±7 x=±14 and x=±2?

Answer 8

\[ 7\sin^2(x)-14\sin(x)+7=0 \]
divide both sides (and all terms) by 7. you get
\[ \sin^2(x)-2\sin(x)+1=0 \]
to avoid confusion, let \( y= \sin x \)
and write this as
\[ y^2 -2 y +1= 0 \]
can you find the values for y ?

Answer 9

I'm so confused. My brother said y=(sqrt-.5)+y

Answer 10

you can factor y^2 -2 y + 1 =0
to get
(y-1)(y-1)=0

now solve for y-1= 0

Answer 11

you get y=1

remember y= sin x
so
sin x =1
what "x" values have sin x = 1?

Answer 12

Letting x = sinx we get \[7x^2-14x+2=-5 \implies 7x^2-14x+7 =0\]
divide both sides by 7 to simplify, we get \[x^2-2x+1=0\] factoring we \[(x-1)^2=0 \implies (x-1)(x-1) = 0\] Now we sub sinx back in to get \[(sinx-1)^2 = 0 \implies (sinx-1)(sinx-1) = 0\] so solving for just sinx -1 = 0, we get sinx = 1

Answer 13

So there's only one value?

Answer 14

sin goes forever.
between 0 and 2pi (or 0 and 360 degrees) there is only pi/2 (90 degrees)
but you can add 2pi to that answer and you will get another value that works
people would write the answer as
\[ x = \frac{\pi}{2} + k \cdot 2 \pi, \text{ k any integer}\]

Answer 15

Thank you so much guys!