Section 7.5 Solving Trigonometric Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

Objectives  Solve trigonometric equations.

Solve Trigonometric Equations When an equation contains a trigonometric expression with a variable, such as cos x, it is called a trigonometric equation. Some trigonometric equations are identities, such as sin 2 x + cos 2 x = 1. Now we consider equations, such as 2 cos x = –1, that are usually not identities. As we have done for other types of equations, we will solve such equations by finding all values for x that make the equation true.

Example Solve First solve for cos x. 2π/3 and 4π/3 have cosine –1/2. Solution: These numbers, plus any multiple of 2π are the solutions.

Example Solve First solve for tan 2x. We are looking for solutions x to the equation for which 0 ≤ x < 2π. Solution: Multiplying by 2, we get 0 ≤ 2x < 4π which is the interval we use when solving tan 2x = –1. Using the unit circle, find the points 2x in [0, 4π) for which tan 2x = –1.

They are: Example (CONT) The values of x, are found by dividing each of these by 2.

Example Solve Solution: Use a calculator in DEGREE mode to find the reference angle = cos – ≈ 65.06º.

Example (cont) Since cos  is positive, the solutions are in quadrants I and IV. The solutions in [0º, 360º) are 65.06º and 360º – 65.06º = º

Example Solve Use the principal of zero products: Solution: The solutions in [0º, 360º) are 60º, 180º and 300º.

Graphical Solution: INTERSECT METHOD Example (cont) and use the INTERSECT feature on the calculator. Graph the equations: The left most solution is 60º. Use the INTERSECT feature two more times to find the solutions, 180º and 300º.

Example (cont) Graphical Solution: ZERO METHOD Write the equation in the form The left most zero is 60º. Use the ZERO feature two more times to find the solutions, 180º and 300º. Then graph

Example Solve Use the quadratic formula: a = 10, b = –12, and c = –7. Solution: No solution. reference angle: 25.44º

Example (cont) Sin x is negative, the solutions are in quadrants III & IV. The solutions are 180º º = º and 300º – 25.44º = º.

Example Solve the following in [0, 2π). Solution: We cannot find the solutions algebraically. We can approximate them with a graphing calculator. On the screen on the next slide, on the left side we use the INTERSECT METHOD. Graph On the screen on the right side we use the ZERO METHOD. Graph

Example (cont) The solutions in [0, 2π) are approximately 1.13 and 5.66.