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= pi/4 ,x=-(3pi)/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{π}{4}, x = - \frac{3 π}{4}

Degrees

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

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{π}{4} + πn

x = \frac{π}{4} + πn

Solutions for the range

- π \leq x \leq π

x = \frac{π}{4}, x = - \frac{3 π}{4}

sin^{2} (x) = 2 - 2 cos (x), 0 \leq x \leq 2 π

tan (θ) = \frac{7}{21}

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

sin (θ - 4 0^{\circ}) = 1

2 cos (x) + 2 sin (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= pi/4 ,x=-(3pi)/4