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Pre-Calc Exam Notes 131

# Pre-Calc Exam Notes 131 - in QI and its re±ection − 3 θ...

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Solving Trigonometric Equations Section 6.1 131 Example6.5 Solve the equation sin θ = tan θ . Solution: Trying the same method as in the previous example, we get sin θ = tan θ sin θ = sin θ cos θ sin θ cos θ = sin θ sin θ cos θ sin θ = 0 sin θ (cos θ 1) = 0 sin θ = 0 or cos θ = 1 θ = 0 , π or θ = 0 θ = 0 , π , plus multiples of 2 π . So since the above angles are multiples of π , and every multiple of 2 π is a multiple of π , we can combine the two answers into one for the general solution: θ = π k for k = 0, ± 1, ± 2, ... Example6.6 Solve the equation cos 3 θ = 1 2 . Solution: The idea here is to solve for 3 θ first, using the most general solution, and then divide that solution by 3. So since cos 1 1 2 = π 3 , there are two possible solutions for 3 θ : 3 θ = π 3 in QI and its
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Unformatted text preview: in QI and its re±ection − 3 θ =− π 3 around the x-axis in QIV. Adding multiples of 2 π to these gives us: 3 θ = ± π 3 + 2 π k for k = 0, ± 1, ± 2, . .. So dividing everything by 3 we get the general solution for θ : θ = ± π 9 + 2 π 3 k for k = 0, ± 1, ± 2, . .. Example 6.7 Solve the equation sin 2 θ = sin θ . Solution: Here we use the double-angle formula for sine: sin 2 θ = sin θ 2 sin θ cos θ = sin θ sin θ (2 cos θ − 1) = ⇒ sin θ = or cos θ = 1 2 ⇒ θ = 0 , π or θ = ± π 3 ⇒ θ = π k and ± π 3 + 2 π k for k = 0, ± 1, ± 2, . .....
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