Question

solve the equation: 2sin^2(x)cos(x)=cos(x) on the interval 0,2pie

Answer 1

it is the only other question i need answers to

Answer 2

solve for \(\bf sin^2(x)\) what would that give you?

Answer 3

hint maybe

Answer 4

hint: linear simplification

Answer 5

i am going into the 12th grade and i dont get what u r saying do you think you can help me a little bit more please

Answer 6

think about of sin(x) = a
and cos(x) = b

so what you really have is -> \(\bf 2a^2b=b\)

Answer 7

if you were to solve for \(\bf a^2\) what would that give?

Answer 8

umh...0/2 maybe

Answer 9

hmmm \(\bf \cfrac{cos(x)}{cos(x)}=0?\)

Answer 10

1

Answer 11

am i wrong

Answer 12

so \(\bf 2sin^2(x)cos(x)=cos(x)\implies 2sin^2(x)=\cfrac{\cancel{ cos(x) }}{\cancel{ cos(x) }}\implies sin^2(x)=\cfrac{1}{2}
\\ \quad \\
\sqrt{sin^2(x)}=\pm\sqrt{\cfrac{1}{2}}\implies sin(x)=\pm\sqrt{\cfrac{1}{2}}
\\ \quad \\
\textit{taking }sin^{-1}\qquad thus
\\ \quad \\
sin^{-1}[sin(x)]=sin^{-1}\left(\pm\sqrt{\cfrac{1}{2}}\right)\implies ?\)

Answer 13

nope.. 1 is correct... so it wouldn't be 0/2, it'd be 1/2

Answer 14

one thing to keep in mind I'd say. is that \(\bf sin^2(x)\to [sin(x)]^2\quad thus \quad \sqrt{sin^2(x)}\implies \sqrt{[sin(x)]^2}\to sin(x)\)

Answer 15

that is....

Answer 16

so.... \(\bf sin^{-1}[sin(x)]=sin^{-1}\left(\pm\sqrt{\cfrac{1}{2}}\right)\implies ?\)

Answer 17

can u maybe give me the answer

Answer 18

well.... skipping material is simple going to create greater issues
so..... "help" here is defined as help you understand it

Answer 19

ok wouldnt the answer be like um a degree

Answer 20

yes... it does.. but you first need to solve for "x"

Answer 21

put it all to x or what

Answer 22

well..... say for example... what's the \(\bf sin(90^o)?\)

Answer 23

\[\sqrt{4}/2\]

Answer 24

hmm welll yes.. and say now what would be \(\bf sin^{-1}\left(\frac{\sqrt{4}}{2}\right)?\)

Answer 25

90 degrees

Answer 26

or pie/2

Answer 27

thanx i think i got it

Answer 28

so... you see.... then that means that \(\bf sin(90^o)={\color{brown}{ \frac{\sqrt{4}}{2}}}\qquad sin^{-1}\left({\color{brown}{ \frac{\sqrt{4}}{2}}}\right)\implies sin^{-1}\left(sin(90^o)\right)\iff 90^o\)