5.5 Solving Trigonometric Equations



In order to solve trigonometric equations, one must understand how to solve the three basic types of trigonometric equations:

I. cos A = xo

II. sin A = yo

III. tan A = m

Type I

To solve cos A = xo, we must find all angles A whose terminal side crosses the unit circle at one of the two points ordered pair x nought comma square root of 1 minus the square of x nought or ordered pair x nought comma negative root of 1 minus x nought squared. It is easier to see this in a diagram of the unit circle. Construct a unit circle and draw the vertical line x = xo. Now it must be the case that interval from minus one to plus one inclusive or the equation will have no solution. The vertical line x = xo will intersect the circle in either one or two places. Any angle A whose terminal side passes through an intersection point of the vertical line and the circle will be a solution to the equation cos A = xo. This is illustrated in the following diagram:

geometric diagram

Example: Solve cos A = - 1 / 2.

Construct the vertical line x = - 1 / 2. It intersects the unit circle at the points ordered pair negative one half comma plus or minus half of root of 3. There are two solutions between 0 and two pi, namely two pi over three and four pi over 3. By adding any multiple of two pi to either of these solutions, we will arrive at other solutions. Since multiples of two pi have the form two pi times n where n is an integer, we say that the general solution to the equation cos x = - 1 / 2 is

A equals 2 pi over 3 plus 2 pi times n or 4 pi over 3 plus 2 pi times n

Exercise 5.5.1

Solve cos x = 0.

Solution

Type II

To solve sin A = yo, we must find all angles A whose terminal side crosses the unit circle at one of the two points ordered pair square root of 1 minus y nought squared comma y nought or ordered pair negative square root of 1 minus y nought squared comma y nought. Construct a unit circle and draw the horizontal line y = yo. Now yo cannot be larger than one nor smaller than negative one or the equation will have no solution. The horizontal line y = yo will intersect the circle in either one or two places. Any angle A whose terminal side passes through an intersection point of the horizontal line and the circle will be a solution to the equation sin A = y. This is illustrated in the following diagram:

Geometric diagram

Example: Solve sin A = 1 / 2.

Construct the horizontal line y = 1 / 2. It intersects the unit circle at the points ordered pair plus or minus half of root of three comma one half. There are two solutions between 0 and 2 pi, namely pi over 6 and 5 pi over 6. Thus, the general solution is
pi over 6 plus two pi times n or 5 pi over 6 plus 2 pi times n

Example: Solve sin A = - 2 / 3.

The horizontal line y = - 2 / 3 intersects the unit circle in the third and fourth quadrants. An angle which passes through the point in the fourth quadrant is arcsin ( - 2 / 3 ), which is the same as
- arcsin ( 2 / 3 ). The angle which passes through the point in the third quadrant is
pi + arcsin ( 2 / 3 ). So the general solution is

minus arcsin of two thirds + 2 pi n or arcsin of two thirds plus pi plus 2 pi n

Exercise 5.5.2

Solve sin A = - 1 / 2

Solution

Exercise 5.5.3

Solve sin A = 0.4.

Solution

Type III

To solve tan A = m, construct the line through the origin having slope m, y = mx. Every angle A in standard position whose terminal side is on the line y = mx is a solution to the equation
tan A = m. In general this will be arctan A plus any multiple of pi, since the period of the tangent graph is pi. So the general solution is A = arctan m + pi n.

Example: Solve tan A = 5.

A = arctan 5 + pi n.

Using these three basic types of trigonometric equations, we can solve other types.

Example: Solve cos2A - cos A = 0.

This can be factored as ( cos A ) ( cos A - 1 ) = 0. Thus, either cos A = 0 or cos A = 1.

The solution to cos A = 0 is pi over 2 plus pi times n. The solution to cos A = 1 is
2 pi times n. So the general solution to the equation is 2 pi n or pi over 2 + 2 pi n.

Exercise 5.5.4

Solve sin2A - sin A = 0.

Solution

Example: Solve cos 2A = half of square root of 3.
The solutions for 2A are 2 A equals plus or minus pi over 6 plus 2 pi n, thus A equals plus or minus pi over twelve plus pi n

Example: Find all solutions in the interval zero to 2 pi, including zero but not including 2 pi for tan ( 3x ) = - 1.

The general solution is three x equals minus pi over 4 plus pi n, that is,

x = minus pi over 12 plus pi times n over 3. Six of these values lie in the interval interval from 0 to 2 pi, including 0 but not 2 pi, namely

minus pi over 12 plus first six multiples of pi over 3, which simplifies to

pi over 4, 7 pi over 12, 11 pi over 12, 5 pi over 4, 19 pi over 12, 23 pi over 12

Exercise 5.5.5

Find all solutions in the interval interval 0 to 2 pi, including 0 but not 2 pi to cos ( 3 x ) = 1 / 2.

Solution

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