# Trigonometry2 sin x tan x + tan x - 2 sin x - 1 = 0 if 0<=x<=pi.

### 2 Answers | Add Yours

You may group the members of equation such that:

`(2 sin x tan x - 2sin x) + (tan x - 1) = 0`

Factoring out `2 sin x` in the first group yields:

`2sin x(tan x - 1) + (tan x - 1) = 0`

Factoring out `(tan x - 1)` yields:

`(tan x - 1)(2sin x + 1) = 0`

Using the zero product property yields:

`{(tan x - 1 = 0),(2sin x + 1 = 0):} => {(tan x = 1),(2sin x = -1):}`

`tan x = 1 => x = pi/4`

`sin x = -1/2` invalid for `0 <= x <= pi.`

**Hence, evaluating the solution to the equation, under the given conditions, yields `x = pi/4 = 45^o` .**

Since tan x = sin x/cos x, we'll re-write the given equation:

2 sin x (sin x/cos x) + sin x/cos x - 2 sin x - 1 = 0

We'll multiply by cos x:

2(sin x)^2 + sin x - 2sin x*cos x - cos x = 0

We'll factorize the first 2 terms by sin x and the last 2 terms by - cos x:

sin x(2 sin x + 1) - cos x(2 sin x + 1) = 0

We'll factorize by 2 sin x + 1:

(2 sin x + 1)(sin x - cos x) = 0

We'll set the first factor as zero:

2 sin x + 1 = 0

We'll subtract 1;

2sinx = -1

sin x = -1/2

x = arcsin (-1/2)

The sine function is negative in the 3rd and 4th quadrants, but not in the given range, therefore the equation sin x = -1/2 has no solutionsover the range (0,pi)

We'll cancel the next factor:

sin x - cos x = 0

This is an homogeneous equation and we'll divide it by cos x:

tan x - 1 = 0

tan x = 1

The function tangent is positive only in the 1st qudrant:

x = arctan 1

x = pi/4 (1st quadrant)

The only solution of the equation, over the range [0 , pi], is: {pi/4}.