Question

solve each equation on the interval [0,2pi)
2 sin theta cos theta = -1

Answer 1

"each"?

Answer 2

do you remember the sine of double angle theorem? :)

Answer 3

*the equation

Answer 4

sin2x = 2sinxcosx

Answer 5

the theorem is \(\sin(2\theta)=2\sin\theta\cos\theta\)

Answer 6

yep

Answer 7

so you can solve it now? :)

Answer 8

but how do i isolate theta?

Answer 9

sin2x = -1

Answer 10

what are the values of 2x?

Answer 11

can i just divide what do you mean?

Answer 12

*divide out the 2

Answer 13

I mean what are the possible values of 2x if sin(2x)=-1? :)

Answer 14

so x is 5pi/6 or pi/6?

Answer 15

nope :/

Answer 16

so how do i get the value of x then?

Answer 17

find the value of 2x first. What will be the range of 2x? :)

Answer 18

-1

Answer 19

the range of x is \([0,2\pi)\), so the range of 2x is?

Answer 20

[0,2pi)?

Answer 21

nvm just find the values of 2x if sin(2x)=-1

Answer 22

sine of what is negative one?

Answer 23

-90

Answer 24

and a few more?

Answer 25

(remember it should be in the interval \([0,4\pi)\))

Answer 26

-pi/2?

Answer 27

\([0,4\pi)\) means that it's greater than 0 and smaller than 4pi

Answer 28

3pi/2

Answer 29

and more?

Answer 30

isn't that it for -1?

Answer 31

what about \(\frac72\pi\)

Answer 32

oh i don't have that on my rad circle

Answer 33

your circle is only to \(2\pi\).
Since x is \([0,2\pi)\), 2x is \([0,4\pi)\).

Answer 34

so now \(2x\) is \(\frac{3\pi}2\) or \(\frac{7\pi}2\)

Answer 35

what would \(x\) be? :)

Answer 36

3pi/4 or 7pi/4

Answer 37

well done :)

Answer 38

thank you soo much!!

Answer 39

no problem :)