Question #85768
2 Answers
Explanation:
We have that
#cos2x = sin2x + 1 - 1#
#cos2x = sin2x#
#cos2x- sin2x= 0#
We now let
#cosu = sinu#
Now consider the
Notice that
the 3rd quadrant is the only quadrant where both sine and cosine are negative, just like the 1st quadrant is the only quadrant where both sine and cosine are positive.
We therefore have that
#u = 2x#
For
#45 = 2x#
#x = 22.5˚#
For
#225 = 2x#
#x = 112.5˚#
However, if we check in the initial equation, we get that
The periodicity of
Our solution set in
Practice exercises
- Solve the following equation for
#x# in the interval#0 ≤ x < 2pi# . Note: the identity#sin^2x + cos^2x = 1# may be useful.
#(cosx - sinx)(cosx + sinx) + 2 = 2sinxcosx + 3#
Solution
#{0˚, 45˚, 135˚, 180˚, 225˚, 315˚}#
Hopefully this helps, and good luck!
22.5; 202.5
112.5; 292.5
Explanation:
Use trig identity:
In this case;
Unit circle gives 2 solutions:
a.
b.
Answers for (0, 360);
22.5; 202.5
112.5; 292.5