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anonymous
 4 years ago
6cos²xsinx4=0
x=?
anonymous
 4 years ago
6cos²xsinx4=0 x=?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i've changed cos²x into 1sin²x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.06cos²xsinx4=0 6(1sin²x)sinx4=0 6sin²xsinx+2=0 (6sin²x+sinx2)=0 then i've used quadratic formula. a=6 b=1 c=2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then i got \[\sin x = (1±\sqrt{1+48})/12\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which means sinx=2/3 or sinx=1/2 when i solved for each of them, didn't give me the correct answer. what did i do wrong?

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.3you can let u=sinx (6u^2+u2)=0 (2u1)(3u+2)=0

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.3u=1/2 u=2/3 so sinx=1/2 sinx=2/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep then sin is positive in QI and QII and sin is negative in QIII and QIV

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so for sinx=2/3, the r.a.a is sin^1(2/3) = 0.73

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which means x = 0.73 or 2.41 since they exist in QI and QII

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and for sinx=1/2, the r.a.a is sin^1(1/2) = π/6 and they exist in QIII and QIV so x = 7π/6 and 11π/6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got final answer as x=0.73, 2.41, 7π/6, and 11π/6 but they're not correct. anything i did wrong there?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2your solution is correct  why do you say its wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the answer sheet says x = π/6, 5π/6, 3.87, 5.55 so I've tried one of my answer, 7π/6 and subbed it in to the equation, and it gave me 1 instead of 0, meaning that my answers are incorrect. :/

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2inverse sin of 0.5 is \(\pi/6\) and \(5\pi/6\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0inverse sin of 0.5 is π/6, but since sinx = 1/2, they should exist in QIII or QIV meaning that x=7π/6 or 11π/6..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sin is negative in QIII or QIV

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2hang on  let me recheck your steps...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok thanks for the help!

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2ah  ok  I see the mistake

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2when you substitute \(\cos^2(x)=1\sin^2(x)\) you made an error in that step. it should have resulted in:\[6\cos^2(x)\sin(x)4=0\]\[6(1\sin^2(x))\sin(x)4=0\]\[66\sin^2(x)\sin(x)4=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hm? 66sin²xsinx4=0 6sin²xsinx+2=0 (6sin²x+sinx2)=0 no? am i still doing something wrong? :/

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2and then:\[6\sin^2(x)\sin(x)+2=0\]\[(2\sin(x)1)(3\sin(x)+2)=0\]giving you:\[\sin(x)=2/3\]or:\[\sin(x)=0.5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh okay. I have no idea why I got the wrong answer for using quadratic formula for that.. somehow I could not factor that. i have no idea why but thanks for the help!
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