Solve equations of the form A·sin(Bx + C) = D, A·cos(Bx + C) = D, or A·tan(Bx) = D. Get all solutions on [0, 2π), the general solution formula, exact π-fraction answers, and a detailed step-by-step walkthrough.
The first step is always to isolate the trig function by dividing both sides by A:
A·sin(Bx + C) = D → sin(Bx + C) = D/A
Once the trig function is alone, apply an inverse trig to get the principal value θ₀.
Because sin and cos are not one-to-one, each principal value gives two angle solutions per period:
sin: θ₀ and π − θ₀
cos: θ₀ and 2π − θ₀ (i.e. −θ₀)
tan: θ₀ only (period π, so θ₀ and θ₀ + π)
Each angle θ is then solved for x: x = (θ − C) / B. Adjust x into [0, 2π) by adding or subtracting multiples of 2π/B.
Solutions on [0, 2π) are just a sample. Because trig functions are periodic, the full solution set is infinite. We express it with an integer parameter n:
sin/cos: x = x₀ + (2π/B)·n or x = x₁ + (2π/B)·n
tan: x = x₀ + (π/B)·n
where n ∈ ℤ (any integer). The solutions in [0, 2π) correspond to specific values of n (usually n = 0 or n = 1).
The period of A·sin(Bx+C) is 2π/B, so each new period adds or subtracts 2π/B to the solutions. For tan the period is π/B.
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