How do you find all solutions for $sin x - tan x = 0$ $sinx - tanx = 0$ where x is between 0 and 2(pi)?

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How do you find all solutions for $sin x - tan x = 0$ $sinx - tanx = 0$ where x is between 0 and 2(pi)?

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Answer 1 · 2015-04-15T15:51:08

$f (x) = sin x - \frac{sin x}{cos x} = sin x \cdot (1 - \frac{1}{cos x}) = sin x \cdot \frac{cos x - 1}{cos x} .$ $f(x) = sin x - sin x/cos x = sin x*(1 - 1/cos x) = sin x*(cos x - 1)/cos x.$
Solve $sin x = 0$ $sin x = 0$ and $(cos x - 1) = 0$ $(cos x - 1) = 0$
Reminder. $f (x)$ $f(x)$ undefined when $cos x = 0 - \to x = \frac{π}{2} and \frac{3 π}{2} .$ $cos x = 0 --> x = pi/2 and (3pi)/2.$
$sin x = 0 - \to x = 0; x = π$ $sin x = 0 --> x = 0; x = pi$ ; and $x = 2 π$ $x = 2pi$
$cos x - 1 = 0 - \to cos x = 1 - \to x = 0; and x = 2 π$ $cos x - 1 = 0 --> cos x = 1 --> x = 0; and x = 2pi$
Answers within period $(0, 2 π) : 0, π, 2 π$ $(0, 2pi): 0, pi, 2pi$ .
Check:
$x = π - \to f (x) = 0 - 0 = 0$ $x = pi --> f(x) = 0 - 0 = 0$ (correct)

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How do you find all solutions for $sin x - tan x = 0$ $sinx - tanx = 0$ where x is between 0 and 2(pi)?

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How do you find all solutions for $sin x - tan x = 0$ $sinx - tanx = 0$ where x is between 0 and 2(pi)?