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

The general solution for sin(x)=-cos(x),-2pi<= x<= 2pi is x=(3pi)/4 ,x=(7pi)/4 ,x=-pi/4 ,x=-(5pi)/4

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sin(x)=-cos(x),-2pi<= x<= 2pi

sin (x) = - cos (x), - 2 π \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = - \frac{π}{4}, x = - \frac{5 π}{4}

Degrees

x = 13 5^{\circ}, x = 31 5^{\circ}, x = - 4 5^{\circ}, x = - 22 5^{\circ}

Solution steps

sin (x) = - cos (x), - 2 π \leq x \leq 2 π

Rewrite using trig identities

sin (x) = - cos (x)

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

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

Solutions for the range

- 2 π \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = - \frac{π}{4}, x = - \frac{5 π}{4}

What is the general solution for sin(x)=-cos(x),-2pi<= x<= 2pi ?
The general solution for sin(x)=-cos(x),-2pi<= x<= 2pi is x=(3pi)/4 ,x=(7pi)/4 ,x=-pi/4 ,x=-(5pi)/4