## anonymous one year ago Find all solutions in the interval [0, 2π). 2 sin2x = sin x

1. Nnesha

set it equal to 0

2. Nnesha

let sin =x to make it easy :D $\huge\rm 2 x^2 = x$set it equal to 0 and then take out common factor

3. anonymous

Can you elaborate more?

4. Nnesha

|dw:1437488293143:dw| alright so first step ) subtract x b oth sides

5. Nnesha

let me know what yo uget

6. anonymous

x^2 = 0

7. Nnesha

no

8. Nnesha

|dw:1437488395098:dw| -x stay same at left side

9. Nnesha

now factor that equation

10. anonymous

X(2^2-1) = 0 X(4-1) = 0 X(3)=0

11. Nnesha

hmm it's x^2 not 2 to the 2 power

12. Nnesha

but yeah x is common factor so take it out $\huge\rm x(2x-1)$ now set x and (2x-1) both equal to zero then solve for x

13. anonymous

x= 1/2 so i would have to look at the unit circle to find the ones with 1/2 as the sine value between 0 and 2π or 0 and 180

14. Nnesha

not between 0 -180 0 -360

15. Nnesha

at what point sin x = 1/2 and sin x = 0 too

16. anonymous

Yes. thank you

17. Nnesha

my pleasure :=)