## anonymous 4 years ago 6cos²x-sinx-4=0 x=?

1. anonymous

i've changed cos²x into 1-sin²x

2. anonymous

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

3. anonymous

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

4. anonymous

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?

5. lalaly

you can let u=sinx -(6u^2+u-2)=0 -(2u-1)(3u+2)=0

6. lalaly

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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?

12. asnaseer

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

13. anonymous

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. :/

14. asnaseer

inverse sin of 0.5 is $$\pi/6$$ and $$5\pi/6$$

15. anonymous

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..

16. anonymous

sin is negative in QIII or QIV

17. asnaseer

hang on - let me recheck your steps...

18. anonymous

ok thanks for the help!

19. asnaseer

ah - ok - I see the mistake

20. asnaseer

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$

21. anonymous

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

22. asnaseer

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$

23. anonymous

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!

24. asnaseer

yw