Math, asked by cshyam678, 1 year ago

the general solution of cosx - sinx equal to 1

Answers

Answered by praneeth1035
0

Answer:

x

=

2

n

π

and

x

=

(

4

n

+

1

)

π

2

,

n

=

0

,

±

1

,

±

2

,

±

3

.

...

Explanation:

The given equation is equivalent to

1

2

sin

x

+

1

2

cos

x

=

1

2

.

This can be written as

cos

(

x

π

4

)

=

cos

(

π

4

)

The general solution of this equation ls

x

π

4

=

2

n

π

±

π

4

,

n

=

0

,

±

1

,

±

2

,

...

,

So,

x

=

2

n

π

and

x

=

(

4

n

+

1

)

π

2

,

n

=

0

,

±

1

,

±

2

,

±

3

.

...

Having noted that there were 40K viewers for the answers by me,

Hero and Nghi, I think I could invoke more interest by including the

solutions for

cos

x

sin

x

=

1

, and for that matter,

sec

x

±

tan

x

=

1

, that become

cos

x

sin

x

=

1

and

cos

x

+

sin

x

=

1

, upon multiplication by

cos x, when

x

an odd multiple of

π

2

.

For cos x - sin x = 1,

the general solution is

x

=

2

n

π

and

x

=

(

4

n

1

)

π

2

,

n

=

0

,

±

1

,

±

2

,

±

3

.

...

Note the change in the multiple from

(

4

n

+

1

)

(

4

n

1

)

.

For

sec

x

±

tan

x

=

1

, it is same sans

(

4

n

±

1

)

π

2

. It is just

x

=

2

n

π

See x-intercepts as graphical solutions.

Graph on uniform scale for solutions

x

=

...

2

π

,

3

π

2

,

0

,

π

2

,

2

π

.

.

of cos x + sin x = 1:

graph{y-cos x - sin x +1 = 0[-7 7 -3 4]}

Graph on uniform scale for solutions

x

=

2

π

,

π

2

,

0

,

3

π

2

,

2

π

,

.

.

of

cos

x

sin

x

=

1

:

graph{y-cos x + sin x +1 = 0[-7 7 -3 4]}

See combined graph for solutions

0

,

±

2

π

,

±

4

π

,

...

of

sec

x

±

tan

x

=

1

:

graph{(y- sec x - tan x +1)(y- sec x+ tan x +1)=0[-13 13 -6.5 6.5]}

Answered by Anonymous
2

Answer:

cos (X) -sin (X) =1

Divide by √2 on both sides

1/√2 cosx-1/√2 sinx=1/√2

therefor ,

➡ cos (π/4) cosx - sin(π/ 4)sinx=cos(π/4)

➡cos(x+π/4)=cos(π/4)

now, costhita =cosalpha implies,

thita =2nπ+ alpha, n€Z

➡x+π/4=2nπ+π/4

➡x=2nπ+π/4- π/4

➡x=2nπ+π/4-π/4 or

X = 2nπ-π/4-π/4

Therefor the general soln is

x=2nπ or x = 2nπ-π/2 , where n€Z

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