Question

I need Help Pleaaase
Find all solutions to the equation.
7 sin2x - 14 sin x + 2 = -5

Answer 1

hmm take common factor to \(\bf 7sin^2(x)-14sin(x)\) what do you get for the common factor there?

Answer 2

2 and 7?

Answer 3

or hmm, lemme do it quick
so \(\bf 7sin^2(x)-14sin(x)+2=-5
\\ \quad \\\
[7sin^2(x)-14sin(x)]=-7
\\ \quad \\
{\color{brown}{ 7sin(x)}}[sin(x)-2]=-7\)

and just solve for sin(x) :)

Answer 4

What do I do to solve for sin(x)

Answer 5

hmmm lemme hmm actually... lemme redo that some

Answer 6

Okay

Answer 7

\(\bf 7sin^2(x)-14sin(x)+2=-5
\\ \quad \\\
[7sin^2(x)-14sin(x)]=-7
\\ \quad \\
{\color{brown}{ 7}}[sin^2(x)-2sin(x)]=-7
\\ \quad \\
sin^2(x)-2sin(x)=\cfrac{\cancel{-7}}{\cancel{7}}
\\ \quad \\
sin^2(x)-2sin(x)=-1\implies {\color{brown}{ sin}}^2(x)-2{\color{brown}{ sin}}(x)+1=0\)

there... notice, it's just a quadratic equation, so, just solve for sin(x), see what you get :)

Answer 8

Is it just pi?

Answer 9

hmm waht did you get for sin(x) anyway?

Answer 10

I got pi because I located point (-1, 0) on the unit circle chart. Is that now how you do it?

Answer 11

not*

Answer 12

hmmm well, I assume you're meant to solve the equation... so , you'd want to solve the quadratic, to see what you get for sin(x) first :)

Answer 13

Okay so
How do I do that?
:(

Answer 14

well, is a quadratic notice \(\begin{array}{llll}
{\color{brown}{ sin}}^2(x)&-2{\color{brown}{ sin}}(x)&+1&=0\\
{\color{brown}{ x}}^2&-2{\color{brown}{ x}}&+1&=0
\end{array}\)

so.. you'd find "the variable" per se, like you'd, on any quadratic
it just so happens that in this case, the variable is sin(x)

Answer 15

for example, if you had say... \(\bf x^2-2x+1=0\) what would you factor that to?

Answer 16

(x-1)^2 + 2(x+1) = 0 ?
Or am I way off

Answer 17

hmm
well, you're supposed to solve for "x" :)

Answer 18

wouldn't it give you a couple of binomials?

Answer 19

oh
yeah
so (x-1)^2=0 and 2(x+1) = 0