What is the general solution for sin(x)+cos(x)=0,-pi<= x<= pi ?

The general solution for sin(x)+cos(x)=0,-pi<= x<= pi is x=(3pi)/4 ,x=-pi/4

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sin(x)+cos(x)=0,-pi<= x<= pi

sin (x) + cos (x) = 0, - π \leq x \leq π

x = \frac{3 π}{4}, x = - \frac{π}{4}

Degrees

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

Solution steps

sin (x) + cos (x) = 0, - π \leq x \leq π

Rewrite using trig identities

sin (x) + cos (x) = 0

Divide both sides by

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

\frac{sin ( x ) + cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

\frac{sin ( x )}{cos ( x )} + 1 = 0

Use the basic trigonometric identity:

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

tan (x) + 1 = 0

tan (x) + 1 = 0

Move

1

to the right side

tan (x) + 1 = 0

Subtract

1

from both sides

tan (x) + 1 - 1 = 0 - 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

Solutions for the range

- π \leq x \leq π

x = \frac{3 π}{4}, x = - \frac{π}{4}

cos (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

2 + 4 cos (θ) = 0

- cos (x) - 4 sin^{2} (x) + 2 = 0

2 sin (t) + cos (2 t) = 0

2 cos^{2} (x) - 1 + cos (x) = 2

What is the general solution for sin(x)+cos(x)=0,-pi<= x<= pi ?
The general solution for sin(x)+cos(x)=0,-pi<= x<= pi is x=(3pi)/4 ,x=-pi/4