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.