Question

maybe a derp question because i just had spring break.. but why is 0 and pi an answer to sin^2x = sin x cos x where u had to give real roots

Answer 1

i got the aswers of pi/4 and 5pi/4 but it also included 0 and pi..

Answer 2

in one of the steps you get 2sin^2x= sin2x
now sin2x and 2sin^2x both will give the answer as zero for pi and 0. and so the LHS and RHS are equal.

Answer 3

i don't think this is a prove the identity question.. at the end i got tanx =1 which = 45 degrees..

Answer 4

hmm \[\large \sin^2x = \sin x \cos x \] since we don't want to remove any solutions by dividing both sides by sinx, maybe start by squaring both sides.

\[\large \sin^4x = \sin^2 x \cos^2 x \]
now use cos^x = 1-sinx^2x
\[\large \sin^4x = \sin^2 x (1- \sin^2 x)\]
simplify a bit...
\[\large \sin^4x = \sin^2 x - \sin^4 x)\]
\[\large \ 2\sin^4x - \sin^2 x =0\]
so
\[\large \sin^2x (2\sin^2x-1) = 0\]

which might give all the required solutions.

Answer 5

\[\large \sin^2 x= 0\]
\[\large 2\sin^2 x - 1= 0\]

Answer 6

for sin^2x = 0, just find where sinx=0 (0, pi)

for the second one:
\[\large \sin x = \frac{ 1 }{ \sqrt2 } = \frac{ \sqrt2 }{ 2}\]

Answer 7

for sin^2x = 0, just find where sinx=0 (0, pi)

for the second one:

\[\large \sin x = \pm \sqrt \frac{ 1 }{ 2 } = \pm \frac{ \sqrt2 }{ 2}\]
@soapia this gives all the required solutions.