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

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

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sin(x)-cos(x)=-2

sin (x) - cos (x) = - 2

No Solution for x \in R

Solution steps

sin (x) - cos (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 the fraction rule:

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

= - \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 (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

2 cos^{2} (x) - sin (x) + 1 = 0, 0 \leq x \leq 18 0^{\circ}

22 cos (x) - 4 sin (x) cos (x) = 0

- sin (x) - 2 sin (2 x) = 0

sin (\frac{π}{6} - x) = 1

t, F = 4 cos (- 2 t + 7) + C

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