Asked by Anonymous
Solve for x: cosx - sinx = 1
Answers
Answered by
oobleck
√2 (1/√2 cosx - 1/√2 sinx) = 1
sin(π/4 - x) = 1/√2
π/4 - x = π/4 or 3π/4
x = 0 or -π/2
in [0,2π] that would be 0 or 3π/2
sin(π/4 - x) = 1/√2
π/4 - x = π/4 or 3π/4
x = 0 or -π/2
in [0,2π] that would be 0 or 3π/2
Answered by
Bosnian
First.
cos ( x + π / 4 )
Use formula:
cos ( a + b ) = cos a ∙ cos b - sin a ∙ sin b
cos ( x + π / 4 ) = cos x ∙ cos π / 4 - sinx ∙ sin π / 4
cos ( x + π / 4 ) = cos x ∙ 1 √2 - sinx ∙ 1 / √2
cos ( x + π / 4 ) = cos x / √2 - sinx / √2
cos x - sin x = 1
Dividie both sides by √2
cos x / √2 - sin x /√2 = 1 / √2
cos ( x + π / 4 ) = 1 / √2
Take inverse cosine of both sides.
Since the period of cos x is 2 π:
x + π / 4 = π / 4 ± 2 π n
Subtract π / 4 from both sides.
x = ± 2 π n
and
x + π / 4 = 2 π n + 7 π / 4
Subtract π / 4 from both sides.
x = 6 π / 4 ± 2 π n
x = 2 ∙ 3 ∙ π / 2 ∙ 2 ± 2 π n
x = 3 / 2 π ± 2 π n
where n = 0 , 1 , 2 , 3...
cos ( x + π / 4 )
Use formula:
cos ( a + b ) = cos a ∙ cos b - sin a ∙ sin b
cos ( x + π / 4 ) = cos x ∙ cos π / 4 - sinx ∙ sin π / 4
cos ( x + π / 4 ) = cos x ∙ 1 √2 - sinx ∙ 1 / √2
cos ( x + π / 4 ) = cos x / √2 - sinx / √2
cos x - sin x = 1
Dividie both sides by √2
cos x / √2 - sin x /√2 = 1 / √2
cos ( x + π / 4 ) = 1 / √2
Take inverse cosine of both sides.
Since the period of cos x is 2 π:
x + π / 4 = π / 4 ± 2 π n
Subtract π / 4 from both sides.
x = ± 2 π n
and
x + π / 4 = 2 π n + 7 π / 4
Subtract π / 4 from both sides.
x = 6 π / 4 ± 2 π n
x = 2 ∙ 3 ∙ π / 2 ∙ 2 ± 2 π n
x = 3 / 2 π ± 2 π n
where n = 0 , 1 , 2 , 3...
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.