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

The general solution for solvefor x,cos(x)=-sin(x) is x=(3pi)/4+pin

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solvefor x,cos(x)=-sin(x)

solve for

x, cos (x) = - sin (x)

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (x) = - sin (x)

Rewrite using trig identities

cos (x) = - sin (x)

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{cos ( x )}{cos ( x )} = \frac{- sin ( x )}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 = - tan (x)

Switch sides

- tan (x) = 1

- tan (x) = 1

Divide both sides by

- 1

- tan (x) = 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{1}{- 1}

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

sin (2 x + 11 0^{\circ}) + sin (2 x - 1 0^{\circ}) = \frac{1}{2}

6 cot^{2} (x) - 18 = 0

(2 sin (θ) - 3) (sin (θ) - 1) = 0

20 sin^{2} (x) + cos (x) - 19 = 0

What is the general solution for solvefor x,cos(x)=-sin(x) ?
The general solution for solvefor x,cos(x)=-sin(x) is x=(3pi)/4+pin