6cos²x-sinx-4=0 x=?

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6cos²x-sinx-4=0 x=?

Mathematics
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i've changed cos²x into 1-sin²x
6cos²x-sinx-4=0 6(1-sin²x)-sinx-4=0 -6sin²x-sinx+2=0 -(6sin²x+sinx-2)=0 then i've used quadratic formula. a=6 b=1 c=-2
then i got \[\sin x = (-1±\sqrt{1+48})/12\]

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which 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?
you can let u=sinx -(6u^2+u-2)=0 -(2u-1)(3u+2)=0
u=-1/2 u=2/3 so sinx=-1/2 sinx=2/3
yep then sin is positive in QI and QII and sin is negative in QIII and QIV
so for sinx=2/3, the r.a.a is sin^-1(2/3) = 0.73
which means x = 0.73 or 2.41 since they exist in QI and QII
and 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
i got final answer as x=0.73, 2.41, 7π/6, and 11π/6 but they're not correct. anything i did wrong there?
your solution is correct - why do you say its wrong?
the 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. :/
inverse sin of 0.5 is \(\pi/6\) and \(5\pi/6\)
inverse 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..
sin is negative in QIII or QIV
hang on - let me recheck your steps...
ok thanks for the help!
ah - ok - I see the mistake
when you substitute \(\cos^2(x)=1-\sin^2(x)\) you made an error in that step. it should have resulted in:\[6\cos^2(x)-\sin(x)-4=0\]\[6(1-\sin^2(x))-\sin(x)-4=0\]\[6-6\sin^2(x)-\sin(x)-4=0\]
hm? 6-6sin²x-sinx-4=0 -6sin²x-sinx+2=0 -(6sin²x+sinx-2)=0 no? am i still doing something wrong? :/
and 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:\[\sin(x)=0.5\]
ohh 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!
yw

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