Solving Trig. Equations
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LammettHash 5 months ago
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In each solution, n is an arbitrary integer.

1. Nothing fancy, just take the inverse tangent.

tan(x) = -5

x = arctan(-5) + nπ

x = -arctan(5) + nπ

2. Recall the identity cos(2x) = 2 cos²(x) - 1, then factorize:

cos(2x) - 3 cos(x) + 2 = 0

(2 cos²(x) - 1) - 3 cos(x) + 2 = 0

2 cos²(x) - 3 cos(x) + 1 = 0

(2 cos(x) - 1) (cos(x) - 1) = 0

2 cos(x) - 1 = 0   or   cos(x) - 1 = 0

cos(x) = 1/2   or   cos(x) = 1

[x = arccos(1/2) + 2nπ   or   x = -arccos(1/2) + 2nπ]

…   or   x = arccos(1) + 2nπ

x = π/3 + 2nπ   or   x = -π/3 + 2nπ   or   x = 2nπ

3. By definition, csc(x) = 1/sin(x) :

csc(2x) + √2 = 0

csc(2x) = -√2

1/sin(2x) = -√2

sin(2x) = -1/√2

2x = arcsin(-1/√2) + 2nπ   or   2x = π - arcsin(-1/√2) + 2nπ

2x = -π/4 + 2nπ   or   2x = 5π/4 + 2nπ

x = -π/8 + nπ   or   x = 5π/8 + nπ

4. More factorization:

2 sin²(x) - sin(x) = 0

sin(x) (2 sin(x) - 1) = 0

sin(x) = 0   or   2 sin(x) - 1 = 0

sin(x) = 0   or   sin(x) = 1/2

x = arcsin(0) + 2nπ

…   or   [x = arcsin(1/2) + 2nπ   or   x = π - arcsin(1/2) + 2nπ]

x = 2nπ   or   x = π/6 + 2nπ   or   x = 5π/6 + 2nπ

5. Yet more factorization. Also recall that |cos(x)| ≤ 1 for all x.

4 cos²(x) - 8 cos(x) + 3 = 0

(2 cos(x) - 3) (2 cos(x) - 1) = 0

2 cos(x) - 3 = 0   or   2 cos(x) - 1 = 0

cos(x) = 3/2   or   cos(x) = 1/2

The first case gives no solution since 3/2 > 1.

x = arccos(1/2) + 2nπ   or   x = -arccos(1/2) + 2nπ

x = π/3 + 2nπ   or   x = -π/3 + 2nπ

6. Recall the identity sin(a - b) = sin(a) cos(b) - cos(a) sin(b).

sin(2x) cos(x) - cos(2x) sin(x) = -√3/2

sin(2x - x) = -√3/2

sin(x) = -√3/2

x = arcsin(-√3/2) + 2nπ   or   x = π - arcsin(-√3/2) + 2nπ

x = -π/3 + 2nπ   or   x = 4π/3 + 2nπ

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