What is the general solution for -2sin(x)-2cos(x)=0 ?

The general solution for -2sin(x)-2cos(x)=0 is x=(3pi)/4+pin

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

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

x = \frac{3 π}{4} + πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

\frac{- 2 sin ( x ) - 2 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

- \frac{2 sin ( x )}{cos ( x )} - 2 = 0

Use the basic trigonometric identity:

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

- 2 tan (x) - 2 = 0

- 2 tan (x) - 2 = 0

Move

2

to the right side

- 2 tan (x) - 2 = 0

Add

2

to both sides

- 2 tan (x) - 2 + 2 = 0 + 2

Simplify

- 2 tan (x) = 2

- 2 tan (x) = 2

Divide both sides by

- 2

- 2 tan (x) = 2

Divide both sides by

- 2

\frac{- 2 tan ( x )}{- 2} = \frac{2}{- 2}

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

7 sin (b) = 5 sin (7 0^{\circ})

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

cos (\frac{1}{2} x) = - 1, y = 2

sin^{2} (a) = 2 sin^{2} (a) + 3

cos (x) = \frac{3}{3}, 0 \leq x \leq 2 π

What is the general solution for -2sin(x)-2cos(x)=0 ?
The general solution for -2sin(x)-2cos(x)=0 is x=(3pi)/4+pin