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anonymous
 one year ago
Find all solutions to the equation.
7 sin2x  14 sin x + 2 = 5
anonymous
 one year ago
Find all solutions to the equation. 7 sin2x  14 sin x + 2 = 5

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mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Not 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I can't factor out 7 because of the 2

phi
 one year ago
Best ResponseYou've already chosen the best response.3first put it in standard form by adding +5 to both sides \[ 7\sin^2(x)14\sin(x)+2=5 \\ 7\sin^2(x)14\sin(x)+7=0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1You can always let sinx = x then factor normally and plug back sinx to find the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry. Are you saying all the possible solutions are x=±7 x=±14 and x=±2?

phi
 one year ago
Best ResponseYou've already chosen the best response.3\[ 7\sin^2(x)14\sin(x)+7=0 \] 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 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm so confused. My brother said y=(sqrt.5)+y

phi
 one year ago
Best ResponseYou've already chosen the best response.3you can factor y^2 2 y + 1 =0 to get (y1)(y1)=0 now solve for y1= 0

phi
 one year ago
Best ResponseYou've already chosen the best response.3you get y=1 remember y= sin x so sin x =1 what "x" values have sin x = 1?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Letting x = sinx we get \[7x^214x+2=5 \implies 7x^214x+7 =0\] divide both sides by 7 to simplify, we get \[x^22x+1=0\] factoring we \[(x1)^2=0 \implies (x1)(x1) = 0\] Now we sub sinx back in to get \[(sinx1)^2 = 0 \implies (sinx1)(sinx1) = 0\] so solving for just sinx 1 = 0, we get sinx = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So there's only one value?

phi
 one year ago
Best ResponseYou've already chosen the best response.3sin 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 \[ x = \frac{\pi}{2} + k \cdot 2 \pi, \text{ k any integer}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much guys!
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