# Find all solutions to the equations: `1. ` `2cos(x)sin(x)-cos(x)=0` 2. sin(x)tan^2(x)=sin(x) 3.tan^2(x)=3

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`2cosxsinx-cosx=0`

To solve, factor out the GCF which is cosx.

`cosx(2sinx-1)=0`

Then, set each factor equal to zero and solve for x.

For the first factor,

`cos x= 0`

Since cosine function is zero when the angle is 90 and 270 degrees, then:

`x= 90^o + 360^o k` and `x=270^o + 360^o k`

Anf for the second factor,

`2sinx-1=0`

Add both sides by 1.

`2sinx+1-1=0+1`

`2sinx=1`

Then, divide both sides by 2.

`(2sinx)/2=1/2`

`sinx=1/2`

Since sine function is equal to 1/2 when the angle is 30 and 150 degrees, then:

`x=30^0+360^o k` and `x=150^0+360^o k`

**Hence, the solutions of the equation 2cosx sinx-cosx=0 are:**

**> `x_1= 30^o + 360^o k ` , **

**> `x_2=90^o + 360^o k` ,**

**> `x_3= 150^o + 360^o k` , and**

**> `x_4=270^0+360^o k` **

**where k is any integer.**

*(For problem #2 and #3, please post it as separate question in Homework Help.)*