How do you solve cos 2x − 2 sin2 x = 0 interval (−5, 5]?

2 Answers
Jun 24, 2018

#-256^@72; -166^@72; -76^@72; 13^@28; 103^@28; 193^@28#

Explanation:

cos 2x - 2sin 2 (1)
x = 0
Divide both sides by cos 2x.
Condition: #cos2x != 0#, --> #2x != pi/2#, and #2x != (3pi)/2#, -->
#x != pi/4#, and #x != (3pi)/4#.
Equation (1) becomes:
1 - 2 tan 2x = 0
#tan 2x = 1/2#
Calculator and unit circle give -->
#2x = 56^@56 + k180^@#.
#x = 13^@28 + k90^@#
The interval (-5, 5) is equivalent to interval #(-286^@62, 286^@62)#
k = 0 --> x = 13.28
k = -1 --> x = 13.28 - 90 = - 76.72 -->
k = -2 --> x = 13.28 - 180 = -16672
k = -3 --> x = 13.28 + 270 = -256.72
k = 1 --> x= 13.28 + 90 = 103.28
k = 2 --> x = 13.28 + 180 = 193.28
The answers for (-5, 5) are:
#-256^@72; -166^@72; -76^@72; 13^@28; 103^@28; 193^@28#

Jun 24, 2018

#x = "Arc""tan" ( 1/2(1-sqrt{5} ) ) pm 45^circ + 180^circ k quad# integer #k#

Explanation:

I'll take the questioner at their word and solve

#cos 2x - 2 sin 2x = 0#

Hopefully they didn't mean #sin^2 x# or #cos ^2 x#

There are a few different ways I can think of to solve this: (A) Let #y=2x# and invert #tan y#. (B) Use the double angle identities, then square to get a solvable quadratic for #cos^2 x.# (C) The linear combination of sine and cosine of the same angle is a scaling and a phase shift.

Let's try (C) first. We convert #(1,-2)# to polar coordinates:

#| (1","-2) |= sqrt{1^2+(-2)^2}=sqrt{5}#

This is fourth quadrant so we can use the principal value of the inverse tangent.

#alpha = angle(1,-2) = "Arc""tan"(-2) approx -63.43°#

We have

#1 = sqrt{5} cos alpha#

#-2 = sqrt{5} sin alpha#

# 1 cos 2x - 2 sin 2x = 0#

# sqrt{5} cos alpha cos 2x + sqrt{5} sin alpha sin 2x = 0#

#sqrt{5} cos(2x - alpha) = 0#

#cos (2x - alpha) = cos (90^circ)#

#2x - alpha = pm 90^circ + 360^circ k#

#x = alpha/2 pm 45^circ + 180^circ k #

#x = 1/2 "Arc""tan"(-2) pm 45^circ + 180^circ k #

The factor of #1/2# on the arctangent is awkward.

We can make progress on

#theta = 1/ 2 alpha# where #alpha = angle(1,-2) = "Arc""tan"(-2) #

We know #alpha#'s in the fourth quadrant so

#cos alpha = 1/sqrt{5}#

#sin alpha = -2/sqrt{5}#

#tan (alpha/2) = sin alpha / {1 + cos alpha} = {1 - cos alpha}/sin alpha#

#tan(alpha/2) = {1 - 1/sqrt{5}}/(-2/sqrt{5}) = 1/2(1-sqrt{5})#

# alpha/2 = 1/2 "Arc""tan"(-2) = "Arc""tan" ( 1/2(1-sqrt{5} ) )#

#x = "Arc""tan" ( 1/2(1-sqrt{5} ) ) pm 45^circ + 180^circ k #

I gotta go so I'll let someone else convert to radians, pull out the answers in range and check it.

~~~~~~~~~~~~~

Attempts at other approaches:

Let #y=2x#

#cos y - 2sin y = 0#

#cos y = 2 sin y#

#tan y = sin y/ cos y= 1/2#

Invoking the multivalued arctangent,

#y = arctan (1/2)#

#2x = arctan(1/2)#

#x = 1/2 arctan (1/2)#

Similar to where we were but simpler.

~~~~~~~~~~~~~~~~~~

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

#2 cos^2 x - 1 = 2 cos x sin x#

Let's square to get square of sines and cosines.

# (2 cos ^2 x-1)^2 = 4 cos^2 x sin ^2 x =4 cos ^2 x(1- cos ^2 x)#

Let #y = cos^2 x#

#(2y -1)^2 = 4y(1-y)#

#4y^2 - 4y + 1 = 4y - 4y^2#

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

# y = 1/8(4 pm sqrt{8}) = 1/4(2 pm sqrt{2})#

Uncle.