Question

Find all values of x in the interval[0,3π] that satisfies the equation sin2x=cosx

Answer 1

start off with sin2x-cosx=0

Answer 2

but we know sin2x=2sinxcosx

Answer 3

therefore we substitute this into our previous equation

Answer 4

2sinxcosx-cosx=0

Answer 5

but there seems to be a common factor. cosx.

Answer 6

cosx(2sinx-1)=0

Answer 7

null factor law says cosx=0 and 2sinx-1=0

Answer 8

hence cosx=0 and sinx=1/2

Answer 9

therefore taking the inverse of both functions...

x=pi/2+k*pi
x=pi/6 +2k*pi
x=5pi/6+2k*pi since sin is positive in quadrant 1 and 2.

Answer 10

since we restrict our domain to the closed interval of 0 and 3pi then our answers are as follows.

Answer 11

x=pi/6,pi/2,5pi/6,3pi/2,13pi/6,5pi/2,17pi/6

Answer 12

we can check this by taking each solution to a couple of decimal places and then graph it and find the roots of the graph sin2x-cox=0