Question

solve each equation for solutions in the interval
0≤x<2π

Answer 1

okay

Answer 2

a) 2sinxcosx=√(3) sinx

Answer 3

2-√(3)=sinx
okay r u agree?

Answer 4

coz sinxcosx=1

Answer 5

\[cos(x)=\frac{\sqrt{3}}{2}\] that's what I get after dividing by sin(x) because it is redundant and dividing by two.

Answer 6

okay why just take sin(x)

Answer 7

this forms seems much easier to me to solve, you can get rid of the sin(x), since it is represented on the lefthand- and also on the righthand side. Therefore it's just another scalar term for the solution.

Remember that
\[2\sin(x)\cos(x)=\sin(2x)\]

Answer 8

yaeh

Answer 9

What you obtain will be:

\[x_1= \frac{\pi}{6} \]
You can get the second solution by looking at the unit circle.

\[x_2= \pi - x_1 = \frac{5 \pi}{6} \]

Answer 10

i ended up getting x=π/3, 4π/3, 3π/4, 7π/4

Answer 11

arm sorry I messed the last term up, it's a full rotation of course, therefore 2 pi

Answer 12

hmm some of your solution don't seem right to me @charliem07, check them by back substitution.

Answer 13

I get

\[x_1 = \frac{\pi}{6} \\
x_2 = 2\pi - x_1 =\frac{11 \pi}{6} \]

Answer 14

and then you can sum them up, using the period.

Answer 15

@Spacelimbus i thought you list them since they are in interval

Answer 16

ohhhh i see what you are saying thanks

Answer 17

you are welcome