Question

sin(2θ)sinθ=cosθ

Answer 1

2sin^2a cosa = cosa
2sin^2a = 1
sina = 1/sqrt2

a = pi/4
cosa= 0
a = pi/2

Answer 2

\[2\cos(\theta)\sin(\theta)\sin(\theta)-\cos(\theta)=0\]
\[\cos(\theta)(2\sin^2(\theta)-1)=0\]

Answer 3

\[\cos(\theta)=0 ,2\sin^2(\theta)-1=0\]

Answer 4

Use:
\[\sin(2x)=2\sin(x)\cos(x)\]
\[2\sin(\phi)\cos(\phi)\sin(\phi)=\cos(\phi)\]
\[2\sin^2(\phi)=1 \implies \sin^2(\phi)=\frac{1}{2} \implies \sin(\phi)=\pm \frac{1}{\sqrt2} \implies \phi=\pm \frac{(2n+1)\pi}{4}\]
\[\cos(\phi)=0 \implies \phi=\pm \frac{(2n+1)\pi}{2}\]

Answer 5

gj ishan but there are more solutions
\[\cos(\theta)=0 => \theta=\frac{\pi}{2}+2n \pi, \theta=\frac{3\pi}{2}+2n \pi\]
\[2\sin^2(\theta)-1=0=>2\sin^2(\theta)=1=>\sin(\theta)=\pm \frac{\sqrt{2}}{2}\]
\[=> \theta=\frac{\pi}{4}+2n \pi, \theta=\frac{3 \pi}{4}+2n \pi, \theta=\frac{5\pi}{4}+2n \pi, \theta=\frac{7\pi}{4}+2n \pi\]

Answer 6

gj male :)

Answer 7

I hope mine is right x.x I'm having a rough day haha

Answer 8

i dont get it so what is the final answer?

Answer 9

look for theta=blah blah
that is the final answer

Answer 10

i my solution i wrote theta=blah blah four different times

Answer 11

Is mine right though myin?

Answer 12

there is infinite amount of solutions

Answer 13

i don't know i didn't read it lol

Answer 14

I mean, just the final result.

Answer 15

i will look

Answer 16

the sin answer looks good
let me look at the cos junk

Answer 17

looks good

Answer 18

gj :)

Answer 19

You too :)

Answer 20

i like how you wrote your together

Answer 21

yours*

Answer 22

instead of separate like mine

Answer 23

It just hit me to write it like that haha

Answer 24

And thank you :DDD I like how we have it both ways though :P

Answer 25

you used one formula instead of 4 for the sin junk
that was nice

Answer 26

It try :)

Answer 27

I*

Answer 28

im still lost guys

Answer 29

Look at my post with all the math, its the one that has the (2n+1)'s in it.

Answer 30

n is an integer

Answer 31

Oh yeah, where n=0,1,2,3,...

Answer 32

n can also be negative integer

Answer 33

n=...,-3,-2,-1,0,1,2,3,...

Answer 34

BUT I HAVE A
\[\pm\]!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 35

or you can just say
\[n \in \mathbb{Z}\]

Answer 36

oh let me see

Answer 37

ok you do but i don't

Answer 38

So you could say
\[n \in \mathbb{Z}^+\]

Answer 39

this is way to complicated i know this isnt what we learned in class it has to do with converting back and forth and some with factoring and stuff

Answer 40

I mean, what we did is the factoring and stuff. The answer must be whats confusing you. There are infinite answers because the trig functions are cyclic. If you have an interval to go by like [0,2pi] then there are finite solutions. Plug in n=0,1,2 into those equations, it gives you ALL the solutions.