# Given is (1 - sin x) * tan x = sin x find all values for x between 0 and 2π

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First of all, we'll substitute tan x= sin x/cos x.

(1 - sin x) * sin x/cos x = sin x

We'll open the paranthesis:

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

We'll write (sin x)^2=1- (cosx)^2

sin x - [1- (cosx)^2]=cos x*sin x

sin x - 1 + (cosx)^2 - cos x*sin x=0

We'll group the terms in this way: the first with the last together, having sin x as common factor. We'll group the middle terms 1- (cos x)^2, seeing the group formed as a difference of squares, type a^2-b^2;

a^2-b^2=(a-b)(a+b)

sin x(1-cosx)-[1-(cosx)^2]=0

sin x(1-cosx)-(1-cosx)(1+cosx)=0

(1-cos x)(sinx-1-cosx)=0

A product of 2 factors is 0, when eather one or the other of the 2 factors is 0.

(1-cos x)=0

cos x=1, **elementary equation**

x=+/- arccos 1 + 2*k*pi

**x=2*pi**, when K=1

**x=0**, when k=0

The second factor, sinx-1-cosx=0

sinx-cosx=1, **linear equation**

This type of equation, due to the fact that the ratio between the coefficient of sin x and cos x is equal to the value 1, which it could be expressed as **tan (pi/4)=1, **could be solved with the help of the known angle. So the equation could be written in this way:

1*sinx-(1*cosx)=1

sin x - tan(pi/4)*cos x=1, but tan (pi/4)=sin(pi/4)/cos (pi/4)

sin x - sin(pi/4)/cos (pi/4)*cos x=1,

We'll have the same denominator all over, by multiplying where necessary with the value cos(pi/4).

sin x*cos (pi/4) - sin(pi/4)*cos x=1*cos (pi/4)

sin [x-(pi/4)]=(sqrt 2)/2, instead of cos(pi/4)=(sqrt 2)/2

x-(pi/4)=(-1)^k*arcsin[(sqrt 2)/2]+k*pi

x=(-1)^k*(pi/4)+(pi/4)+k*pi

When k=0

x=(pi/4)+(pi/4)

x=2*(pi/4)

**x=pi/2**

When k=1

x=-(pi/4)+(pi/4)+pi

**x=pi**

(1-sinx)tanx =sinx

(1-sinx)sinx= sinxcosx

1-sinx = cosx

1-sinx = sqrt(1-sin^2)

(1-sinx)^2=1-(sinx)^2

1-2sinx+(sinx)^2 = 1-(sinx)^2

2(sinx)^2 - 2sinx = 0

2sinx(sinx-1) = 0

sonx=0 or sinx = 1

sinx =0 for x= 0 degree and x=180 degree.

sinx =1 for x= 90 degree.

After going through it I still don't know what I did that broke the equality of the problem in my previous comment, but I have a decently fast way to the solution.

(1 - sin x) tan x = sin x

divide both sides by tan x

1 - sin x = cos x

use the trig id cos^2 x + sin ^2 x = 1 substituting 1 - sin x for cos

(1 - sin x)^2+ sin ^2 x = 1

expand (1 - sin x)^2

sin^2 x - 2 sin x + 1 + sin ^2 x = 1

add like terms and subtract 1 from both sides

2sin^2 x - 2 sin x = 0

factor 2sin x

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

treat it like a polynomial and find the points where it is 0

sin x = 0 at x = 0

sin x - 1 = 0

sin x = 1 at x = pi/2

I am very out of practice with these, and while I know i didn't take anything close to a direct route, I'm not quite sure quite where I went wrong.

(1-sin(x))tan(x) = sin(x)

multiply both sides by (1+sin(x)) as 1-sin^2(x) = cos^2(x)

(1+sin(x))(1-sin(x))tan(x) = sin(x)(1+sin(x))

use the trig id (1+sin(x))(1-sin(x)) = 1-sin^2(x) = cos^2(x)

cos^2(x)tan(x) = sin(x)(1+sin(x))

divide both sides by cos(x) remembering sin(x)/cos(x) = tan(x)

cos(x)tan(x) = tan(x)(1+sin(x))

tan(x) cancels

cos(x) = 1+sin(x)

again 1 - sin^2(x) = cos^2(x) = (1+sin(x))(1-sin(x))

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

substituting for 1+sin(x)

cos(x) = cos^2(x)/(1-sin(x))

cos(x) cancels

1 = cos(x)/(1-sin(x))

after multiplying by (1-sin(x))

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

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

cos^2(x)+sin^2(x) = 1 which is true for all values of x