# Solve for x the equation cos x-(tan x*cos x)=0, 0<=x=<2pi

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cos(x) - tan(x)*cos(x) = 0

**Making use of the fact that tan(x) = sin(x)/cos(x), we have:**

cos(x) - [sin(x)/cos(x)]/cos(x) = 0

cos(x) - sin(x) = 0

**Moving sin(x) to the other side, we can see that:**

cos(x) = sin(x)

**Now ask yourself, where on the unit circle is cos(x) = sin(x)?**

**This occurs only at pi/4 and 5pi/4.**

*Note that at pi/2 and 3pi/2, the tan is undefined.*

**Therefore, the solution set is: {pi/4, 5pi/4}.**

We have to solve cos x-(tan x*cos x)=0 for 0<=x=<2*pi

cos x-(tan x*cos x)=0

=> cos x(1 - tan x) = 0

cos x = 0

=> x = pi/2 and x = 3*pi/2

1 - tan x = 0

=> tan x = 1

=> x = pi/4

**The solution of the equation is x = pi/2, 3*pi/2 and pi/4**

We'll factorize by cos x:

cos x(1 - tan x) = 0

We'll cancel each factor:

cos x = 0

x = arccos 0

x = `pi` /2 or x = 3`pi`

We'll cancel the next factor:

1 - tan x = 0 => tan x = 1

x = arctan 1

x = `pi` /4 (1st quadrant) or x = `pi` + `pi` /4 = 5`pi` /4 (3rd quadrant)

**The solutions of the equation are: {`pi`/4 ; `pi`/2 ; 5`pi` /4 ; 3`pi` /2 }.**