What is the general solution for sin(x)+cos(x)=-3 ?

The general solution for sin(x)+cos(x)=-3 is No Solution for x\in\mathbb{R}

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sin(x)+cos(x)=-3

sin (x) + cos (x) = - 3

No Solution for x \in R

Solution steps

sin (x) + cos (x) = - 3

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}) = - 3

Divide both sides by

2

2 sin (x + \frac{π}{4}) = - 3

Divide both sides by

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{- 3}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{- 3}{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{- 3}{2} : - \frac{3 2}{2}

\frac{- 3}{2}

Apply the fraction rule:

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

= - \frac{3}{2}

Rationalize

- \frac{3}{2} : - \frac{3 2}{2}

- \frac{3}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{3 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{3 2}{2}

= - \frac{3 2}{2}

sin (x + \frac{π}{4}) = - \frac{3 2}{2}

sin (x + \frac{π}{4}) = - \frac{3 2}{2}

sin (x + \frac{π}{4}) = - \frac{3 2}{2}

- 1 \leq sin (x) \leq 1

No Solution for x \in R

sin (x) = \frac{1}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{2 5}{5}

csc (x) = cot (x) - 1

2 cos (θ) + 1 = sec (θ)

tan (θ) = - \frac{2}{2}

What is the general solution for sin(x)+cos(x)=-3 ?
The general solution for sin(x)+cos(x)=-3 is No Solution for x\in\mathbb{R}