## jpjones Group Title solve the following equation for x where 0<or = to x<2pi, sin2x=cosx 2 years ago 2 years ago

1. Mimi_x3

hint: $$sin2x = 2sinxcosx$$

2. jpjones

yes and

3. Mimi_x3

well where are you stuck or show me what you did

4. jpjones

the equation is not solved

5. Mimi_x3

$2sinxcosx = cosx => 2sinxcosx - cosx = 0$ why dont you try and solve it

6. sami-21

take cos(x) you will have $\Large \cos(x)[2\sin(x)-1]=0$ now you have two equations $\Large \implies \cos(x)=0$ $\Large \implies 2\sin(x)-1=0$ solve these two equations.

7. jpjones

if i knew how to solve these equations i wouldn;t be asking for help

8. jpjones

to this stage i know, but how to solve from here?

9. Mimi_x3

$\cos(x) = 0 => cosx = \cos\left(\frac{\pi}{2}\right) => x = \frac{\pi}{2} +n*2\pi , n \epsilon\mathbb{Z}$ $sinx = \sin\left(\frac{\pi}{6}\right) => x = \left(-1\right)^{n}*\frac{\pi}{6} +n*\pi , n \epsilon\mathbb{Z}$

10. jpjones

i've been to wolfram alpha to mimi

11. Mimi_x3

lol i didnt use wolfram..

12. jpjones

the answer is correct though but i have to show it using the unit circle, or the graph

13. jpjones

i think using the sin wave

14. Mimi_x3

|dw:1345619287591:dw|

15. sami-21

ok let me give you an example then if you have 2cos(x)+1=0 then $\Large \implies \cos(x)=\frac{-1}{2}$ $\Large x=\cos^{-1}(\frac{-1}{2})$ there are only two points in 9o.2pi) for which cos(x) is -1/2 they are $\Large x=\frac{2 \pi}{3}$ $\Large x=\frac{4 \pi}{3}$ now solve the 2sin(x)-1=0 its quite similar to this.

16. Mimi_x3

for sinx |dw:1345619404355:dw|

17. sami-21

you use wolfram,thats why you are unable to solve these equations by yourself. this wolfram thingy has ruined the students :(. i think at this level students should avoid using that.

18. Mimi_x3

yeah i agree. you should only use it to check your answers

19. jpjones

no i dont use it but i am aware of it

20. jpjones

mimi you've been great i think i should become a fan