What is the general solution for cos(x)+sin(x)=2 ?

The general solution for cos(x)+sin(x)=2 is No Solution for x\in\mathbb{R}

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cos(x)+sin(x)=2

cos (x) + sin (x) = 2

No Solution for x \in R

Solution steps

cos (x) + sin (x) = 2

Rewrite using trig identities

sin (x) + cos (x)

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

Rewrite as

= 2 (\frac{1}{2} sin (x) + \frac{1}{2} cos (x))

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= 2 sin (x + \frac{π}{4})

2 sin (x + \frac{π}{4}) = 2

Divide both sides by

2

2 sin (x + \frac{π}{4}) = 2

Divide both sides by

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{2}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{2}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (x + \frac{π}{4})

Simplify

\frac{2}{2} : 2

\frac{2}{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{1}}{2 ^{\frac{1}{2}}} = 2^{1 - \frac{1}{2}}

= 2^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 2^{\frac{1}{2}}

Apply radical rule:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= 2

sin (x + \frac{π}{4}) = 2

sin (x + \frac{π}{4}) = 2

sin (x + \frac{π}{4}) = 2

- 1 \leq sin (x) \leq 1

No Solution for x \in R

sin (3 θ) cos (θ) - cos (3 θ) sin (θ) = \frac{3}{2}

tan (x) = 3 cot (x)

2 cos (\frac{1}{2} x) - 1 = 0

2 cos (θ) sin (θ) - sin (θ) = 0

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

What is the general solution for cos(x)+sin(x)=2 ?
The general solution for cos(x)+sin(x)=2 is No Solution for x\in\mathbb{R}