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

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

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sin(z)+cos(z)=2

sin (z) + cos (z) = 2

No Solution for z \in R

Solution steps

sin (z) + cos (z) = 2

Rewrite using trig identities

sin (z) + cos (z)

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

sin (z) + cos (z)

Rewrite as

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

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 (z) + sin (\frac{π}{4}) cos (z))

Use the Angle Sum identity:

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

= 2 sin (z + \frac{π}{4})

= 2 sin (z + \frac{π}{4})

2 sin (z + \frac{π}{4}) = 2

Divide both sides by

2

2 sin (z + \frac{π}{4}) = 2

Divide both sides by

2

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

Simplify

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

Simplify

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

\frac{2 sin ( z + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (z + \frac{π}{4})

Simplify

\frac{2}{2} : 2

\frac{2}{2}

Apply radical rule:

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:

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

= 2

sin (z + \frac{π}{4}) = 2

sin (z + \frac{π}{4}) = 2

sin (z + \frac{π}{4}) = 2

- 1 \leq sin (x) \leq 1

No Solution for z \in R

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- 3 tan (θ) - 11 = 3 tan (θ) - 6

cos^{2} (x) - 6 cos (x) - 1 = 0

tan (5 x) tan (x) = tan (5 x)

4 sin^{2} (x) = 1 + 4 cos (x)

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