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# HW10sol - MAC1114 Homework 10 Due Tuesday March 22 • For...

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Unformatted text preview: MAC1114 - Homework 10 Due Tuesday - March 22 • For each of the following equations: (a) Find all solutions. (b) Solve in 1. 3 tan3 x = tan x. (a) 3 tan3 x−tan x = 0 ⇒ tan x(3 tan2 x−1) = 0 ⇒ (b) Plug in 2. x = 0, 1 to get tan x = 0 ⇒ 3 tan2 x − 1 = 0 tan x = 0 √ tan x = ± 33 x = nπ, n ∈ Z ⇒ x = π + nπ, n ∈ Z . 6 π x = 56 + nπ, n ∈ Z π π {0, π , 56 π, 76 , 11π }. 6 6 sec2 4x − 1 = 3. (a) 4x 4x sec2 4x = 4 ⇒ sec 4x = ±2 ⇒ cos 4x = ± 1 ⇒ 2 4x 4x (b) Plug in 3. [0, 2π ). x = 0, 1, 2, 3 to get π π { 12 , 5π , π , π , 7π , 11π , 23 , 12 6 3 12 12 π π x = 12 + nπ 3 + 2nπ 2 x = 5π + nπ 5π + 2nπ 3 12 2 ⇒ 2π x = π + nπ 3 + 2nπ 6 2 4π x = π + nπ 3 + 2nπ 3 2 5π 13π 17π 7π 4π 19π 23π 5π 11π , 12 , 12 , 6 , 3 , 12 , 12 , 3 , 6 }. 6 = = = = 2 sin x + csc x = 0 1 1 2 sin x = − csc x ⇒ 2 sin x = − sin x ⇒ sin2 x = − 2 . This has no solution since the left-hand side is always positive. 4. 2 sin2 x − 7 sin x + 3 = 0 (a) Set u = sin x (ignore part (b) for this one) to get 2u2 −7u+3 = 0 ⇒ (2u−1)(u−3) = 0 ⇒ 2 sin x − 1 = 0 ⇒ sin x = 3 1 sin x = 1 2 ⇒ x = π + 2nπ, x = 11π + 2mπ. 6 6 sin x = 3 ...
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