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

##### 2 Answers

#### Explanation:

cos 2x - 2sin 2 (1)

x = 0

Divide both sides by cos 2x.

Condition:

Equation (1) becomes:

1 - 2 tan 2x = 0

Calculator and unit circle give -->

The interval (-5, 5) is equivalent to interval

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:

#### Explanation:

I'll take the questioner at their word and solve

Hopefully they didn't mean

There are a few different ways I can think of to solve this: (A) Let

Let's try (C) first. We convert

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

We have

The factor of

We can make progress on

We know

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

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Attempts at other approaches:

Let

Invoking the multivalued arctangent,

Similar to where we were but simpler.

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

Let

Uncle.