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Mathematics Skill Development - Module 7 Mathematics Skill Development - Module 7 Trigonometric Equations The following questions will evaluate the student’s ability to solve various trigonometric equations. 1. Solve the following equation sin x = − sin2 x. Solution: Rewriting the equation as sin x = − sin2 x sin2 x + sin x = 0 sin x(1 + sin x) = 0, we have that its solutions correspond to sin x = 0 or sin x = −1. In the first case, sin x = 0 when x = kπ for any integer k ∈ Z. Meanwhile, sin x = −1 when x = 3π 2 + 2kπ for any integer k ∈ Z. Therefore, there is an infinite amount of number x satisfying this equation characterized by x = kπ and x = 3π 2 + 2kπ where k could be any integer. 2. Solve the following tan x 4 = 1. Solution: Since tan(π/4) = 1 and tan(y + π) = tan(y) for any number y, the solutions are given by x π = + kπ, k ∈ Z. 4 4 Therefore, there is an infinite number of solutions to this equation characterized by x = π + 4kπ where k could be any integer. 3. Solve the following sin x cos x = 1 . 2 Solution: We recall the trigonometric identity sin 2x = 2 sin x cos x, to rewrite the left hand side of this equation as Therefore, this equation is equivalent to sin(2x) = 1. Since the solutions to sin y = 1 are given by y = π/2 + 2kπ where k ∈ Z, we obtain 2x = π/2 + 2kπ for any integer k or sin(2x) . 2 x= π + kπ, 4 k ∈ Z. 1 Mathematics Skill Development - Module 7 4. Solve the following tan2 x cos x + 2 sin x + cos x = 0. Solution: Dividing this equation by cos x, we obtain a quadratic equation for tan x: 2 sin x +1=0 cos x 2 tan x + 2 tan x + 1 = 0 tan2 x + (tan x + 1)2 = 0. Therefore tan x = −1 or x = −π/4 + kπ for any integer k. Since we divided by cos x in the first step, it is important to check that the solution we obtain does not correspond to cos x = 0. Since cos π4 = √12 and √1 cos 3π 4 = − 2 , this is not the case for any of the solutions we obtained. 5. Solve the following 2 cos x − sin x + 2 cos x sin x = 1. Solution: We simplify the equation by factoring it 2 cos x − sin x + 2 cos x sin x = 1 2 cos x sin x + 2 cos x − sin x − 1 = 0 2 cos x(sin x + 1) − (sin x + 1) − 0 (sin x + 1)(2 cos x − 1) = 0. Therefore sin x = −1 or cos x = 21 . In the first case, sin x = −1 if x = x = π3 + 2kπ or x = 5π 3 + 2kπ. 3π 2 + 2kπ. Meanwhile, cos x = 1 2 if 2