Question

Find all solutions in the interval [0,2(pi)]
sin2x=cosx

Answer 1

You can rewrite sin2x using the double angle formula to get:

cosx = 2*sinx*cosx... subtract cosx on both sides
2sinxcosx - cosx = 0... factor
cosx(2sinx - 1) = 0

Set each factor equal to zero to get two sets of roots:
cosx = 0 ==>
x = {π/2,3π/2}
2sinx - 1 = 0 ==> sinx = 1/2 ==>
x = {π/6, 5π/6}

Answer 2

\[\sin(2x)=2\sin(x)\cos(x)\] so put
\[2\sin(x)\cos(x)=\cos(x)\]
\[2\sin(x)\cos(x)-\cos(x)=0\]
\[\cos(x)(2\sin(x)-1)=0\]
\[\cos(x)=0\] or
\[2\sin(x)-1=0\]

Answer 3

damn i am slow

Answer 4

slowpoke

Answer 5

:)

Answer 6

lol. but i am using latex!

Answer 7

stop making excuses