Trigonometric Equations Advanced

[MathDefinition title="Trigonometric Equations"] A trigonometric equation is an equation containing one or more trigonometric functions of unknown angles. The general solution gives all possible values of the unknown angle. [/MathDefinition]

[MathDefinition title="General Solutions"] The general solutions for basic trigonometric equations are:

1. If sin θ = 0, then θ = nπ, n ∈ Z
2. If cos θ = 0, then θ = (2n+1)π/2, n ∈ Z
3. If tan θ = 0, then θ = nπ, n ∈ Z
4. If sin θ = sin α, then θ = nπ + (−1)ⁿ α, n ∈ Z
5. If cos θ = cos α, then θ = 2nπ ± α, n ∈ Z
6. If tan θ = tan α, then θ = nπ + α, n ∈ Z [/MathDefinition]

[MathKeyFormula title="Key Formulas for General Solutions"]

1. sin θ = sin α ⟹ θ = nπ + (−1)ⁿ α
2. cos θ = cos α ⟹ θ = 2nπ ± α
3. tan θ = tan α ⟹ θ = nπ + α
4. sin² θ = sin² α ⟹ θ = nπ ± α
5. cos² θ = cos² α ⟹ θ = nπ ± α
6. tan² θ = tan² α ⟹ θ = nπ ± α [/MathKeyFormula]

[MathNote title="Solving Strategy"]

1. Reduce the equation to a single trigonometric function
2. Express the equation in the form sin θ = k, cos θ = k, or tan θ = k
3. Find the principal solution
4. Write the general solution using appropriate formula
5. Verify solutions in the given domain if specified [/MathNote]

[MathExample title="Example 1: Solving sin θ = 1/2"] Problem: Find the general solution of sin θ = 1/2.

Solution: We know that sin(π/6) = 1/2

Using the formula: if sin θ = sin α, then θ = nπ + (−1)ⁿ α

Here α = π/6

General solution: θ = nπ + (−1)ⁿ (π/6), n ∈ Z

Verification: For n = 0: θ = π/6 ✓ For n = 1: θ = π − π/6 = 5π/6 ✓ For n = 2: θ = 2π + π/6 = 13π/6 ✓ [/MathExample]

[MathExample title="Example 2: Solving cos 2θ = cos θ"] Problem: Find the general solution of cos 2θ = cos θ.

Solution: cos 2θ = cos θ

Using cos 2θ = 2cos² θ − 1: 2cos² θ − 1 = cos θ 2cos² θ − cos θ − 1 = 0

Let x = cos θ: 2x² − x − 1 = 0 (2x + 1)(x − 1) = 0

Therefore: x = 1 or x = −1/2

Case 1: cos θ = 1 θ = 2nπ, n ∈ Z

Case 2: cos θ = −1/2 cos θ = cos(2π/3) θ = 2nπ ± 2π/3, n ∈ Z

General solution: θ = 2nπ or θ = 2nπ ± 2π/3, n ∈ Z [/MathExample]

[MathExample title="Example 3: Solving tan θ = tan 2θ"] Problem: Find the general solution of tan θ = tan 2θ.

Solution: tan θ = tan 2θ

Using the identity: if tan A = tan B, then A = nπ + B

θ = nπ + 2θ −θ = nπ θ = −nπ = nπ (since −n is also an integer)

But we must exclude values where tan is undefined. tan θ is undefined when θ = (2m+1)π/2

Therefore, the general solution is: θ = nπ, n ∈ Z, where n is not a half-integer.

Let's verify by substituting back: tan(nπ) = 0, tan(2nπ) = 0 ✓

Actually, let me reconsider. If tan θ = tan 2θ, we need to consider the domain.

From tan θ = tan 2θ: sin θ/cos θ = sin 2θ/cos 2θ sin θ cos 2θ = cos θ sin 2θ sin θ cos 2θ − cos θ sin 2θ = 0 sin(θ − 2θ) = 0 sin(−θ) = 0 −sin θ = 0 sin θ = 0 θ = nπ, n ∈ Z

This is valid as long as tan θ and tan 2θ are defined, which they are when θ = nπ. [/MathExample]

[MathExample title="Example 4: Solving sin² θ = sin² 2θ"] Problem: Find the general solution of sin² θ = sin² 2θ.

Solution: sin² θ = sin² 2θ sin² θ − sin² 2θ = 0

Using the identity sin² A − sin² B = sin(A+B) sin(A−B): sin(θ + 2θ) sin(θ − 2θ) = 0 sin 3θ sin(−θ) = 0 −sin 3θ sin θ = 0 sin 3θ sin θ = 0

Therefore: sin 3θ = 0 or sin θ = 0

Case 1: sin θ = 0 θ = nπ, n ∈ Z

Case 2: sin 3θ = 0 3θ = nπ θ = nπ/3, n ∈ Z

The general solution is: θ = nπ/3, n ∈ Z (this includes both cases) [/MathExample]