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Pre-Calc Exam Notes 129

# Pre-Calc Exam Notes 129 - .” since it will be understood...

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6 Additional Topics 6.1 Solving Trigonometric Equations An equation involving trigonometric functions is called a trigonometric equation . For exam- ple, an equation like tan A = 0.75 , which we encountered in Chapter 1, is a trigonometric equation. In Chapter 1 we were concerned only with ±nding a single solution (say, between 0 and 90 ). In this section we will be concerned with ±nding the most general solution to such equations. To see what that means, take the above equation tan A = 0.75. Using the a tan 1 calculator button in degree mode, we get A = 36.87 . However, we know that the tangent function has period π rad = 180 , i.e. it repeats every 180 . Thus, there are many other possible answers for the value of A , namely 36.87 + 180 , 36.87 180 , 36.87 + 360 , 36.87 360 , 36.87 + 540 , etc. We can write this in a more compact form: A = 36.87 + 180 k for k = 0, ± 1, ± 2, . .. This is the most general solution to the equation. Often we will simply leave out the part that says “for k = 0, ± 1, ± 2, .
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Unformatted text preview: ..”, since it will be understood that k varies over all integers. The general solution in radians would be: A = 0.6435 + π k for k = 0, ± 1, ± 2, . .. Example 6.1 Solve the equation 2 sin θ + 1 = 0. Solution: Isolating sin θ gives sin θ = − 1 2 . Using the ✄ ✂ a ✁ sin − 1 calculator button in degree mode gives us θ =− 30 ◦ , which is in QIV. Recall that the re²ection of this angle around the y-axis into QIII also has the same sine. That is, sin 210 ◦ = − 1 2 . Thus, since the sine function has period 2 π rad = 360 ◦ , and since − 30 ◦ does not differ from 210 ◦ by an integer multiple of 360 ◦ , the general solution is: θ = − 30 ◦ + 360 ◦ k and 210 ◦ + 360 ◦ k for k = 0, ± 1, ± 2, . .. In radians, the solution is: θ = − π 6 + 2 π k and 7 π 6 + 2 π k for k = 0, ± 1, ± 2, . .. For the rest of this section we will write our solutions in radians. 129...
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