Solve the following equation for theta where 0 < or= theta < 2 pie sin ^2 theta = sin theta
\[\sin^2(\theta)-\sin(\theta)=0\] factor left hand side
is that (sin theta) (sin theta)???
yes x^2 can be written as x*x
so... now what?
factor
\[x^2-x \\ x(x-1)\] you try factoring \[\sin^2(\theta)-\sin(\theta)\] it is the same way as I did the expression of x
\[x^2=x \\ x^2-x=0 \\ x(x-1)=0 \\ \text{ set both factors equal to zero } \\ x=0 \text{ or } x-1=0 \\ x=0 \text{ or } x=1 \]
you try solving for sin(theta) now it is the same way
once you have solved for sin(theta) then you will easily be able to find theta
hold on, don't leave yet
sin =0 sin theta = 0
I have no idea what to do next
well you have one equation correct but you are missing an equation
have you not factored \[\sin^2(\theta)-\sin(\theta)\] yet?
do you see both sin^2(theta) and sin(theta) have a common factor?
as you said sin^2(theta) can be written as sin(theta)*sin(theta) therefore sin(theta) and sin^2(theta) both have the common factor sin(theta)
so you can factor that out of the expression
\[\sin^2(\theta)-\sin(\theta)=0 \\ \sin(\theta)(\sin(\theta)-1)=0\] that is what you should get for the left hand side after factoring now set both factors equal to 0
you can use the unit circle to solve both equations
wait hold on
where did you get these 2 equations?
which two?
wait nvm
ok but how do you use the unit circle for this? i'm lost
do you know which equations you are solving?
sin^2 theta - sin theta = 0
you need to factor the left hand side
I have actually done this for you
I know x factor, but don't know how
|dw:1433565703539:dw|
now set both factors equal to 0
you have sin(theta)=0 or sin(theta)-1=0
find when sin(theta) is 0 on the unit circle find when sin(theta) is 1 on the unit circle
in other words you are looking for where the y-coordinate is 0 on the unit circle and also looking for when the y coordinate is 1 on the unit circle
the answer is pie/2 and pie
but how is it pie though???
nvm
got it
so you don't get that sin(pi)=0? also sin(0)=0 so I think you mean you have 0,pi/2,pi
ok thx
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