Many trigonometric equations are more complex than the ones seen so far. Some of these can be solved using many of the algebraic strategies you learned in previous courses. The key is to isolate the trigonometric term. This is often done by collecting like terms, as in the following example.
Read “Example 1” on pages 207 and 208 of the textbook.
Share 2 is based on “Example 1.”
With a partner or group, discuss how solving 5 sin θ + 2 = 1 + 3 sin θ compares to solving the equation 5x + 2 = 1 + 3x.
If required, save a record of your discussion in your course folder.
Complete questions 1.a., 1.c., 3.a., 3.b., 4.c., and 4.e. on page 211 of the textbook. Answer
So far you have solved first-degree trigonometric equations. Next you will explore how to solve a second-degree trigonometric equation by factoring. A second-degree trigonometric equation includes a trigonometric ratio squared, such as (sin x)2 = 0.45. Often (sin x)2 is written as sin2x.
The solutions to a trigonometric equation can be checked by substitution, just like other equations. To substitute, replace the variable with a solution and check that the equation is true.
Take a look at Solving Trigonometric Equations 2. Notice the similarities between solving a trigonometric equation by factoring and solving a quadratic equation by factoring.
