Answer:
As a general description, there are 3 steps. These steps may be very challenging, or even impossible, depending on the equation.
Step 1: Find the trigonometric values need to be to solve the equation.
Step 2: Find all 'angles' that give us these values from step 1.
Step 3: Find the values of the unknown that will result in angles that we got in step 2.
(Long) Example
Solve:
2sin(4x−π3)=1
Step 1: The only trig function in this equation is
sin
.
Sometimes it is helpful to make things look simpler by replacing, like this:
Replace sin(4x−π3) by the single letter S. Now we need to find S to make 2S=1. Simple! Make S=12
So a solution will need to make sin(4x−π3)=12
Step 2: The 'angle' in this equation is (4x−π3). For the moment, let's call that θ. We need sin θ=12
There are infinitely many such θ, we need to find them all.
Every θ that makes sinθ=12 is coterminal with either π6 or with 5π6. (Go through one period of the graph, or once around the unit circle.)
So θ Which, remember is our short way of writing 4x−π3 must be of the form: θ=π6+2πk for some integer k or of the form θ=5π6+2πk for some integer k.
Step 3:
Replacing θ in the last bit of step 2, we see that we need one of: 4x−π3=π6+2πk for integer k or 4x−π3=5π6+2πk for integer k.
Adding π3 in the form 2π6 to both sides of these equations gives us: 4x=3π6+2πk=π2+2πk for integer k or 4x=7π6+2πk for integer k.
Dividing by 4 (multiplying by 14) gets us to: x=π8+2πk4 or x=7π24+2πk4 for integer k.
We can write this in simpler form:
x=π8+π2k or x=7π24+π2k for integer k.
Final note: The Integer
k could be a positive or negative whole number or 0. If k is negative, we're actually subtracting from the basic solution.
