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Math 111, Trigonometric Equation Problems
Darin Comeau, Section 4
Here are some additional problems where you solve trigonometric equations. For each of the below equations, find all x that make the statement true. Refer to the online notes for some worked examples.
- sin x tan x = sin x, x ∈ [0, 2 π]
Solution: Subtracting sin x from both sides, we have
sin(x) tan x − sin x = 0 sin x(tan x − 1) = 0
⇒ sin x = 0 or tan x − 1 = 0 In [0, 2 π], sin x = 0 for x = 0, π, 2 π, and tan x = 1 for x = π/ 4 , 5 π/4 (remember tan x is positive for x terminating in quadrants I and III). Then our solutions are x = 0,
π 4
, π,
5 π 4
, 2 π
♠
- tan^2 x + tan x − 2 = 0
Solution: We recognize this as a quadratic in the variable tan x. Then using the quadratic formula, we get the roots of the above equation are
tan x =
Therefore the roots of the quadratic equation are tan x = − 2 , tan x = 1. These are the equations we now need to solve, but we’ll first write the quadratic in its factored form to see why we are solving them.
tan^2 x + tan x − 2 = (tan x − 1)(tan x + 2) = 0
Since tan x is giving us our solutions, we will find all solutions in (−π/ 2 , π/2) (there will be one from each equation) and add integer multiples of π, the period of tangent. tan x = 1 is true for x = π/4, and tan x = −2, taking arctan of both sides gives us x = arctan(−2). Note in this example we only have to look for one solution to each equation since tan x is one-to-one on one cycle, here (−π/ 2 , π/2). Then we add πn to each solution to find all solutions:
x =
{ (^) π 4 +^ πn arctan(−2) + πn
- cos(2x) = cos x
Solution: We want to use a double angle identity for cos 2x that will leave us with factors of cos x; so we use cos 2x = 2 cos^2 x − 1. Substituting this in, we have: 2 cos^2 x − 1 = cos x 2 cos^2 x − cos x − 1 = 0
We now factor using the quadratic formula; we have
cos x =
Therefore the solutions to the quadratic equation (note these are not solutions to our problem) are cos x = 1, − 1 /2:
2 cos^2 x − cos x − 1 = (cos x − 1)
cos x +
So we first find the solutions to cos x = 1 in [0, 2 π), which is only x = 0 (since this is the maximum of cos x. The solutions to cos x = − 1 /2 in [0, 2 π) are x = 2π/ 3 , 4 π/3. Then adding 2πn to all our solutions, we have
x =
0 + 2πn 2 π 43 π^ + 2πn 3 + 2πn ♠
- 4 sin x cos x =
Solution: We recognize the identity sin(2x) = 2 sin x cos x, and use this to rewrite the expression in terms of sin(2x). We first divide each side by 2:
2 sin x cos x =
sin(2x) =
Now we know how to find the solutions to sin θ =
3 /2; the solutions in [0, 2 π) are θ = π/ 3 , 2 π/3. So letting θ = 2x, we get x = π/ 6 , π/3. Now here is where we would normally add 2πn to each solution; however here the function that is giving us our solutions is sin(2x), not sin x, and the period of sin(2x) is π. So we actually need to add integer multiples of π instead:
x =
{ (^) π (^6) π +^ πn 3 +^ πn
