http://www.ck12.org Chapter 6. Analytic Trigonometry
x=^196 π,^236 π
- Start by factoring:
2 cos^2 x+3 cosx− 2 = 0
(2 cosx− 1 )(cosx+ 2 ) = 0Note that cosx 6 =−2 which means only one equation needs to be solved for solutions.
2 cosx− 1 = 0
cosx=^12
x=π 3 ,−π 3These are the solutions within the interval−πtoπ. Since this represents one full period of cosine, the rest of the
solutions are just multiples of 2πadded and subtracted to these two values.
x=±π 3 ±n· 2 πwherenis an integer
- There are an infinite number of possible equations that will work. When you see theπ 4 you should think either of
where tangent is equal to one or where sine/cosine is equal to
√ 2
- The problem with both of these initial guesses
is that tangent repeats everyπnot every 2π, and sine/cosine have a second place where they reach a height of 1. An
option that works is:
tanx 2 = 1
This equation works because the period of tanx 2 is 2π.
Practice
Solve each equation on the interval[ 0 , 2 π).
- 3 cos^2 x 2 = 3
- 4 sin^2 x=8 sin^2 x 2
Find approximate solutions to each equation on the interval[ 0 , 2 π). - 3 cos^2 x+10 cosx+ 2 = 0
- sin^2 x+3 sinx= 5
- tan^2 x+tanx= 3
- cot^2 x+5 tanx+ 14 = 0
- sin^2 x+cos^2 x= 1
Solve each equation on the interval[ 0 , 360 ◦). - 2 sin(x−π 2 )= 1
- 4 cos(x−π) = 4
Solve each equation on the interval[ 2 π, 4 π). - cos^2 x+2 cosx+ 1 = 0