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 yaxis 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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 Fall '11
 Dr.Cheun
 Calculus, Trigonometry, Addition, Equations, general solution, single solution

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