SolvTrigEq

SolvTrigEq - Solving Trig Equations Section 1 In algebra...

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Solving Trig Equations Section 1 In algebra class you spent a lot of time learning to solve equations. For example 2x + 1 = 15 which has the solution x = 7, and x 2 +1 = 10 which has two solutions x = 3 and x = -3. Keep in mind that we are finding values of x that will satisfy the equation. In this unit we will learn to solve equations where the unknown is “inside” a trig function. For example we will learn to solve sin(2x + 1) = 0. The key to solving equations like this is to use the graph of sine or cosine to get started. The figure to the right shows the graph of y = sin(x). When we solve the equation sin(x) = 0 we are finding values of x for which sin(x) is zero. On the graph this means values of x where the y-coordinate is zero. We don’t have to do any work. All we do is read the x values from the graph. Therefore six solutions to sin(x) = 0 are x = -2 π , – π , 0 , π , 2 π , 3 π and we could continue this pattern to get as many solutions as desired. Suppose we want to solve the equation cos(x) = 1. We would look at the graph of y=cos(x) and read the x values that have a y coordinate equal to 1. So the four solutions are x = -2 π , 0, 2 π , 4 π and infinitely many more following the same pattern. Exercises : 1. Give three solutions to sin(x) = 1. 2. Give three solutions to cos(x) = 0. 3. Give five solutions to sin(x) = -1. 4. Give four solutions to cos(x) = -1. Most of the trig equations we will encounter in applications are not quite as simple as those shown above. But solving those will be the basis for solving more complicated equations. For example suppose we want to solve the equation sin(2x – 1) = 0. The key
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2 here is to recognize that the equation has the form sine of “something” equals 0, or in symbols sin( θ ) = 0. We can use the graph of sine to solve sin( θ ) = 0 and get θ = 0 or θ = π or θ = 2 π or θ = 3 π etc. In the original equation sin(2x – 1) = 0 we have θ = 2x – 1. Therefore from the line above we can write 2x – 1 = 0 2x – 1 = π 2x – 1 = 2 π 2x –1 = 3 π which we can solve to get 2x = 1 2x = π + 1 2x = 2 π + 1 2x = 3 π + 1 x = 1 2 x = π + 1 2 x = 2 π + 1 2 x = 3 π + 1 2 We believe that these are solutions to sin(2x – 1) = 0. Recall from algebra that you can check a solution by substituting the value back into the equation and see if the equation is satisfied. For example to check the solution x = 1 2 above, we would substitute 1 2 for x in on the left side of the equation sin(2x – 1) = 0 and we get sin(2( 1 2 ) 1) = sin(1 = sin(0) = 0 which means that x = 1 2 is a solution. To check x = π + 1 2 we would calculate sin(2( π + 1 2 ) = sin(( π + = sin( π ) = 0 . Another example : Find three positive solutions to cos( π x) = 1.
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This note was uploaded on 10/18/2011 for the course MAC 1114 taught by Professor Russel during the Fall '08 term at Valencia.

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SolvTrigEq - Solving Trig Equations Section 1 In algebra...

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