How do you solve $sin x - tan x = 0$ $sinx-tanx=0$ from [0,2pi]?

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How do you solve $sin x - tan x = 0$ $sinx-tanx=0$ from [0,2pi]?

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Answer 1 · 2015-07-27T22:15:34

I found: $x = 0, π, 2 π$ $x=0,pi,2pi$

Explanation:

Write it as:
$sin (x) - \frac{sin (x)}{cos (x)} = 0$ $sin(x)-sin(x)/cos(x)=0$ collect $sin (x)$ $sin(x)$
$sin (x) [1 - \frac{1}{cos (x)}] = 0$ $sin(x)[1-1/cos(x)]=0$
So you get:
$sin (x) = 0$ $sin(x)=0$ for $x = 0, π, 2 π$ $x=0, pi, 2pi$
$1 - \frac{1}{cos (x)} = 0$ $1-1/cos(x)=0$ or $cos (x) = 1$ $cos(x)=1$ for $x = 0, 2 π$ $x=0, 2pi$

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How do you solve $sin x - tan x = 0$ $sinx-tanx=0$ from [0,2pi]?

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How do you solve sinx-tanx=0sinx−tanx=0 sinx-tanx=0 from [0,2pi]?

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How do you solve $sin x - tan x = 0$ $sinx-tanx=0$ from [0,2pi]?