Question

Solve the equation cos x + sin x = 1

Answer 1

Solution:
➖➖➖➖
on multiplying both side by 1/√2 .

$cos \: x \times \frac{1}{ \sqrt{2} } + \sin(x) \times \frac{1}{ \sqrt{2} } = \frac{1}{ \sqrt{2} } \\ \\ \cos(x) \cos(45°) + \sin(x) \sin(45°) = \frac{1}{ \sqrt{2} }$

now we are applying the formula of
cos a cos b + sin a sin b

$\cos(A) \cos(B) +\sin(A) \sin(B) = \cos(A - B)$

$\cos(x) \cos(45°) +\sin(x) \sin(45°) = \frac{1}{ \sqrt{2} } \\ \\ \cos(x - 45°) = \frac{1}{ \sqrt{2} } \\ \\ x - 45° = {cos}^{ - 1}( \frac{1}{ \sqrt{2} } ) \\ \\ x - 45° = {cos}^{ - 1}( \cos(45°) ) \\ \\ x - 45° = 45° \\ \\ x = 45° + 45° \\ \\ x = 90°\\\\ x = \frac{π}{2}$

Answer 2

Answer: x=2nπ

x=2nπ+π/2

Step-by-step explanation:

cos x + sin x= 1

=>1/√2cosx+1/√2sinx=1/√2

=>cosxcosπ/4+sinxsinπ/4=cosπ/4

=>cos(x-π/4)=cosπ/4

=>cos(x-π/4)-cosπ/4=0

=>2sin((x-π/4+π/4)/2)sin((x-π/4-π/4)=0

=>2sin(x/2)sin(x/2-π/4)=0

=>sin(x/2)sin(x/2-π/4)=sin(nπ)

=>sin(x/2)=sin(nπ) ----1

=>sin(x/2-π/4)=sin(nπ)----2

where n belongs to Int.

Therefore,

x/2=nπ

x=2nπ ✓✓

x/2-π/4=nπ

x=2nπ+π/2✓✓

FORMULAE USED

1) acosx+bsinx=c

2) cos(A-B)=cosAcosB+sinAsinB

3) 2sin((C+D)/2)sin((C-D)/2) = cosC - cosD