Question

6cos²x-sinx-4=0
x=?

Answer 1

i've changed cos²x into 1-sin²x

Answer 2

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

Answer 3

then i got \[\sin x = (-1±\sqrt{1+48})/12\]

Answer 4

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?

Answer 5

you can let u=sinx

-(6u^2+u-2)=0
-(2u-1)(3u+2)=0

Answer 6

u=-1/2
u=2/3

so sinx=-1/2
sinx=2/3

Answer 7

yep then sin is positive in QI and QII
and sin is negative in QIII and QIV

Answer 8

so for sinx=2/3, the r.a.a is sin^-1(2/3) = 0.73

Answer 9

which means x = 0.73 or 2.41 since they exist in QI and QII

Answer 10

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

Answer 11

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?

Answer 12

your solution is correct - why do you say its wrong?

Answer 13

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. :/

Answer 14

inverse sin of 0.5 is \(\pi/6\) and \(5\pi/6\)

Answer 15

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..

Answer 16

sin is negative in QIII or QIV

Answer 17

hang on - let me recheck your steps...

Answer 18

ok thanks for the help!

Answer 19

ah - ok - I see the mistake

Answer 20

when you substitute \(\cos^2(x)=1-\sin^2(x)\) you made an error in that step. it should have resulted in:\[6\cos^2(x)-\sin(x)-4=0\]\[6(1-\sin^2(x))-\sin(x)-4=0\]\[6-6\sin^2(x)-\sin(x)-4=0\]

Answer 21

hm?
6-6sin²x-sinx-4=0
-6sin²x-sinx+2=0
-(6sin²x+sinx-2)=0
no? am i still doing something wrong? :/

Answer 22

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:\[\sin(x)=0.5\]

Answer 23

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!

Answer 24

yw