Question #0f638

This equation is equivalent to

#sqrt(sin^2x) = sqrt(cos^2(3x))#

now squaring both sides

#sin^2x = cos^2(3x) rArr (sinx +cos3x)(sinx-cos3x) = 0#

and now knowing that

#cos3x = cos^3x-3cosx sin^2x#

#{(sinx+cos^3x-3cosx sin^2x = 0),(sinx-cos^3x+3cosx sin^2x = 0):}#

Those equations to be completely solved require the solution of a cubic polynomial. We left this as an exercise.

We have #abs(sinx)=abs(cos3x)#. But by definition, #cos3x=sin(pi/2-3x)#.

So the equation becomes #abs(sinx)=abs(sin(pi/2-3x))#.

This equation is true iff #sinx=sin(pi/2-3x)# or #sinx=-sin(pi/2-3x)=sin(3x-pi/2)#

Finding general solutions will be left as an exercise

#|sinx|=|cos3x|#

From the tables we know that :

#color(blue)(cos3x=4 cos^3 x-3 cosx)#,

#|sinx|=|4 cos^3 x-3 cosx|#

#+-sinx= (4cos^2 x-3)cosx#

#+-tan x= (4cos^2 x-3)#

#color(blue)(1+tan^2 x=1/cos^2x)#

#tan^2 x=( (4cos^2 x-3))^2=(16cos^4x-24cos^2 x+9)#

#1/cos^2x=16cos^4x-24cos^2 x+9+1#

#1=16cos^6x-24cos^4 x+10Cos^2x#

I could solve this equation with #y =cos^2x#

#16y^3-24y^2+10y-1=0#

but it is beyond my capabilities.

If we play a little with the trigonometric circle we can find

that #x=+-pi/4+kpi# is one of the solutions, as

#|sin(x)|=|sin(pi/4)+kpi|=sqrt(2)/2#

and

#|cos(3x)|=|cos(3pi/4)+3kpi|=sqrt(2)/2#

#color(red)(x=pi/4+kpi/2#

so #x=pi/4-> cos(x)=sqrt(2)/2-> y=0.5#

We just need to divide the third degree equation by (y-0.5) and obtain the remainder solutions:

#16y^2-16y+2=0#

#8y^2-8y+1=0#

#y=(8+-sqrt(8^2-4*8*1))/(2*8)#

#y=(8+-sqrt(64-32))/(16)#

#y=(8+-sqrt(32))/(16)#

#y=(8+-4sqrt(2))/(16)#

#y=(2+-sqrt(2))/(4)#

#->cos^2x=y#

#cosx=sqrt((2+-sqrt(2))/(4)#

#cosx=sqrt(2+-sqrt(2))/(2)#

#cosx=cos(+-pi/8) or cos x=cos (+-(3pi)/8)#

#x=+-pi/8+2kpi or x +-3pi/8+2kpi#

#color(red)(x=pi/8 + kpi/2)#

Remember the first solution:

#color(red)(x=pi/4+kpi/2#

and gives

#color(green)(x=pi/8 + (kpi)/4)#

Note that sin x = cos (pi/2 - x)
#Icos (pi/2 - x)I = Icos 3xI#
We have to solve these equations:
#cos (pi/2 - x) = cos 3x#, and
#cos (pi/2 - x) = - cos 3x#
A. #cos (pi/2 - x) = cos 3x#
#(pi/2 - x) = +- 3x#
a. #pi/2 - x = 3x# --> #4x = pi/2 + 2kpi# --> #x = pi/8 + (kpi)/4#
b. #pi/2 - x = - 3x# --> #2x = - pi/2 + 2kpi# --> #x = - pi/4 + kpi#
B. #cos (pi/2 - x) = - cos 3x = cos (3x + pi)#
#(pi/2 - x) = +- (3x + pi)#
a. #pi/2 - x = 3x + pi# --> #4x = - pi/2 + 2kpi# -->
#x = - pi/4 + kpi#
b. #pi/2 - x = - 3x - pi# --> #2x = - (3pi)/2 + 2kpi# -->
#x = - ((3pi)/4) + kpi#
Finally, the general solutions are:
#x = pi/8 + (kpi)/4; x = - pi/4 + kpi; x = -((3pi)/4) + kpi#
Check.
#x = pi/8 = 22^@5# --> #sin 22.5 = 0.38# -->
#cos 67.5 = 0.38#. Proved
#x = - 45# --> #sin (- 45) = - sqrt2/2# -->
#cos (- 135) = - sqrt2/2#. Proved
#x = - (3pi)/4# --> #sin x = - sqrt2/2# -->
#cos ((9pi)/4) = cos (pi/4) = sqrt2/2#. Proved.