Solving Trigonometric

EquationsTrig identities are true for all values of the variable for which the variable is defined.

However, trig equations, like algebraic equations, are true for some but not all values of the variable.

Trig equations do not have unique solutions.Trig equations have infinitely many solutions.They differ by the period of the function, 2 or

360° for sin and cos, and or 180° for tan.

Solving Trig Equations

Trig equations will have unique solutions if the value of the function is restricted to two adjacent quadrants.

These solutions are called the principal values.For sin x and tan x, the principal values are in

Quadrants I or IV. x is in the interval -90° < x < 90°.

For cos x, the principal values are in Q I or II.So x is in the interval 0° < x < 180°.

Solve 2 cos² x – 5 cos x + 2 = 0 for the principal values of x.

2 cos² x – 5 cos x + 2 = 0

(2 cos x - 1) (cos x – 2) = 0 Factor

2 cos x - 1 = 0

2 cos x = 1

cos x = ½

x = 60°

or cos x – 2 = 0

cos x = 2

There is no solution for cos x = 2 since –1 < cos x < 1

Solve 2 tan x sin x + 2 sin x = tan x + 1 for all values of x.

2 tan x sin x + 2 sin x = tan x + 1

2 tan x sin x + 2 sin x – tan x – 1 = 0 Subtract tan x + 1 from both sides.

(tan x + 1) (2 sin x – 1) = 0 factor

tan x + 1 = 0

tan x = -1

or 2 sin x – 1 = 02 sin x = 1sin x = ½

When all the valuesof x are required, the solution shouldbe represented asx + 360k° for sin and cos, and x + 180k° for tan,where k is anyinteger.

x = -45° + 180k°x = 30° + 360k°

or

x = 150° + 360k°

Solve sin² x + cos 2x – cos x = 0 for the principal values of x.

sin² x + cos 2x – cos x = 0

sin² x + (1 – 2 sin² x) – cos x = 0Express cos 2x in terms

of sin x.

1 – sin² x – cos x = 0 Combine like terms

cos² x – cos x = 0 1 – sin² x = cos² x

cos x (cos x – 1) = 0 Factor

cos x = 0

x = 90°

or cos x – 1 = 0cos x = 1x = 0°

So the solutions are 0° and 90°

Solve cos x = 1 + sin x for 0° < x < 360°

cos x = 1 + sin xcos² x = (1 + sin x)² Square both sides

cos² x = 1 + 2 sin x + sin² x Expand the binomial squared

1 – sin² x = 1 + 2 sin x + sin² x cos² x = 1 – sin² x

0 = 2 sin x + 2 sin² x

0 = 2 sin x (1 + sin x) Factor

2 sin x = 0

sin x = 0x = 0° or 180º

or 1 + sin x = 0sin x = -1

x = 270º

It’s important to always check your

solutions. Some may not actually be solutions to the original equation.

x = 0º or 180º x = 270º

cos x = 1 + sin xcos 0º = 1 + sin 0°

1 = 1cos 180º = 1 + sin 180º-1 = 1 + 0-1 ≠ 1

☻

cos 270º = 1 + sin 270º0 = 1 + (-1)0 = 0 ☻

Based on the check, 180º is not a solution

Assignment

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