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.
~~~~~~~~~~~~~
Attempts at other approaches:
Let
Invoking the multivalued arctangent,
Similar to where we were but simpler.
~~~~~~~~~~~~~~~~~~
Let's square to get square of sines and cosines.
Let
Uncle.