Question

solve the equation 2[(sin^2 x)(cosx)]=cosx by factoring. give the solutions on [0, 2pi]

Answer 1

\[2(\sin ^{2}x)(cosx)=cosx\]

Answer 2

\(\bf 2[sin^2(x)cos(x)] = cos(x) \implies sin^2(x) = \cfrac{\cancel{cos(x)}}{2\cancel{cos(x)}}\\
sin(x) = \sqrt{\cfrac{1}{2}} \implies sin(x) = \cfrac{1}{\sqrt{2}} \implies sin(x) =\cfrac{\sqrt{2}}{2}\)

Answer 3

\(\bf sin(x) = \pm \cfrac{\sqrt{2}}{2} \implies sin^{-1}(sin(x)) = sin^{-1} \left(\pm \cfrac{\sqrt{2}}{2}\right)\)

Answer 4

Subtracting cosine from each side,\[\Large 2\sin^2x \cos x-\cos x=0\]Factoring a cosine out of each term,\[\Large \cos x(2\sin^2x-1)=0\]

Doe's method works also, but we end up losing a solution because of the cancellation of those cosines :c

Answer 5

hmm, I see that

Answer 6

ok, so with either way how do you find the other solutions, or is there only one? i forget how the renaming the angles works on the unit circle

Answer 7

renaming the angles? :o

Answer 8

i dont know what its called but isnt there something about having to change them to fit in the right quadrant or something im not really sure

Answer 9

\[\Large \cos x(2\sin^2x-1)=0\]Using this method, we would set each factor equal to zero and solve for the trig functions separately.\[\Large \cos x=0 \qquad\qquad\qquad 2\sin^2x-1=0\]

Where is cosx=0?
Recall that cosine is the x component of our angle.
The x-coordinate is 0 when we're up at pi/2 and down at 3pi/2.
So those are 2 of our solutions.
\(\large x=\pi/2\) and \(\large x=3\pi/2\)

Answer 10

right quadrant? Hmm we might have to worry about that when we deal with the sine factor, let's see...

Answer 11

\[\Large 2\sin^2x-1=0\]Solving for sin^2x gives us,\[\Large \sin^2x=\frac{1}{2}\]And then solving for sinx, taking the root of both sides gives us,\[\Large \sin x=\pm\frac{\sqrt{2}}{2}\]Hmm

Answer 12

ok so the \[+\frac{ \sqrt{2} }{ 2 } \] would work, but the negative one wouldnt because it isnt within 2pi so would there only be those three answers?

Answer 13

or would pi/4 just be the answer for the second part

Answer 14

The negatives ARE in 2pi, right? :)
I was just trying to think about whether or not they would actually work for a solution.
Since the sinx is being squared, it looks like all 4 angles will work.

\[\large x=\pi/4, \qquad x=3\pi/4, \qquad x=5\pi/4\qquad x=7\pi/4\]

Lemme test one just to make sure I'm not being silly.

Answer 15

\[\large x=5\pi/4 \qquad\to\qquad 2\sin^2(5\pi/4)\cos(5\pi/4)=\cos(5\pi/4)\]

Soooo this gives us,\[\large 2\left(-\frac{\sqrt2}{2}\right)^2\left(-\frac{\sqrt2}{2}\right)=\left(-\frac{\sqrt2}{2}\right)\]Which simplifies to,\[\large 2\left(\frac{1}{2}\right)\left(-\frac{\sqrt2}{2}\right)=\left(-\frac{\sqrt2}{2}\right)\]

Yah that seems to work ok :)

Answer 16

A little rusty on your trig angles maybe? :D

Answer 17

haha yeah :)

Answer 18

Bahhh not the best drawing.. i was trying to show that the sine represents the vertical (y component) of that angle.
So you can clearly see when the y value is negative for those 2 lower angles.

Answer 19

Ehh maybe that's more confusing XD lol.
Just brush up on your trig!! :3

Answer 20

so there are six answers to this question?

Answer 21

Here maybe this will help :)
https://www.desmos.com/calculator/fq8u4tetb1

Answer 22

Solutions means ~ points of intersection.
How many times do they intersect between 0 and 2pi?

I tried to label the x-axis in radians, so it'd be easier to read.
Hopefully that feature is showing up correctly.

Answer 23

ok thank you so much for all of your help with this!!

Answer 24

np c:

I find trig to be a little bit difficult to explain without going into a ton of detail.
So maybe just look back at a unit circle or something.
I hope you can make sense of those 6 solutions :[