How do you solve #mcosx+msinx=cos2x#?

2 Answers
Cesareo R.
Aug 8, 2017

See below.

Explanation:

As we know

#cos(2x)=cosx cosx-sinx sinx# then

#m = cosx = -sinx# or

#tanx = -1# or

#x = -pi/4 + k pi# with #k=0,pm1,pm2,pm3,cdots#

Nghi N
Aug 9, 2017

#x = (3pi)/4#
#x = (7pi)/4#
#m = +- sqrt2#

Explanation:

#m(cos x + sin x) = cos^2 x - sin^2 x = (cos x - sin x)(cos x + sin x)#
Put (cos x + sin x) in common factor:
#(cos + sin x )(m - (cos x - sin x)) = 0#
Either factor should be zero.
a. #cos x + sin x = sqrt2cos (x - pi/4) = 0#
#cos (x - pi/4) = 0#
a. #(x - pi/4) = pi/2# --> #x = pi/2 + pi/4 = (3pi)/4#
b. #(x - pi/4) = (3pi)/2# --> #x = (3pi)/2 + pi/4 = (7pi)/4#

To find m, solve the equation:
m - (cos x - sin x) = 0
#m = cos x - sin x = sqrt2cos (x + pi/4)#
Replace x by the 2 above values.
c. #m = sqrt2cos ((3pi)/4 + pi/4) = sqrt2cos (pi) = - sqrt2#
d. #m = sqrt2cos ((7pi)/4 + pi/4) = cos (2pi) = sqrt2#
#m = +- sqrt2#
Check by calculator
#x = (3pi)/4 = 135^@#. Compute the given equation:
#mcos x + msin x = cos 2x#
#sqrt2(-sqrt2/2) + sqrt2(sqrt2/2) = cos 270 = 0#. Proved

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