Find the exact solution of each equation over the interval [0,2pi)???

sin(x) = cos(2x)

1 Answer
Aviv S.
Mar 17, 2018

a) #x=pi/6,(5pi)/6,(3pi)/2#

b) #x=pi/4,(3pi)/4, (5pi)/4, (7pi)/4#

c) #x=pi/9, (5pi)/9, (7pi)/9, (11pi)/9, (13pi)/9, (17pi)/9#

d) #x=pi#

Explanation:

Here's problem a):

#sinx=cos2x#

#sinx=1-2sin^2x#

#2sin^2x+sinx-1=0#

#(2sinx-1)(sinx+1)=0#

#sinx=1/2,sinx=-1#

Here's a unit circle to remind us of some sine and cosine values:

enter image source here

#x=pi/6,(5pi)/6,(3pi)/2#

Here's problem b):

#4sin^2x + 2cos^2x= 3#

#4sin^2x+2(1-sin^2x)=3#

#4sin^2x+2-2sin^2x=3#

#2sin^2x-1=0#

#sin^2x-1/2=0#

#(sinx+sqrt(1/2))(sinx-sqrt(1/2))=0#

#sinx=+-sqrt(1/2)#

#color(white)(sinx)=+-sqrt1/sqrt2#

#color(white)(sinx)=+-1/sqrt2#

#color(white)(sinx)=+-sqrt2/2#

Using the unit circle from before:

#x=pi/4,(3pi)/4, (5pi)/4, (7pi)/4#

Here's problem c):

#2cos3x=1#

#cos3x=1/2#

#3x=pi/3, (5pi)/3, (7pi)/3, (11pi)/3, (13pi)/3, (17pi)/3...#

#x=pi/9, (5pi)/9, (7pi)/9, (11pi)/9, (13pi)/9, (17pi)/9#

Here's problem d):

#2sin(x/3) - sqrt(3) = 0#

#2sin(x/3)=sqrt3#

#sin(x/3)=sqrt3/2#

#x/3=pi/3, (2pi)/3#

#x=pi,color(red)cancelcolor(black)(2pi)qquadlarrqquad"outside of "[0,2pi)#

Those are all the answers. I hope this helped!

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