What is the general solution for 4cos(x)-4sin(x)=0 ?

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

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4cos(x)-4sin(x)=0

4 cos (x) - 4 sin (x) = 0

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

4 cos (x) - 4 sin (x) = 0

Rewrite using trig identities

4 cos (x) - 4 sin (x) = 0

Divide both sides by

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

\frac{4 cos ( x ) - 4 sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

4 - \frac{4 sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

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

4 - 4 tan (x) = 0

4 - 4 tan (x) = 0

Move

4

to the right side

4 - 4 tan (x) = 0

Subtract

4

from both sides

4 - 4 tan (x) - 4 = 0 - 4

Simplify

- 4 tan (x) = - 4

- 4 tan (x) = - 4

Divide both sides by

- 4

- 4 tan (x) = - 4

Divide both sides by

- 4

\frac{- 4 tan ( x )}{- 4} = \frac{- 4}{- 4}

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

3 csc (3 x) = - 2

cos (4 α) - cos (2 α) = sin (3 α)

3 sec (2 x) + 2 = 0, 0 \leq x < 2 π

1 sin (x) tan (x) - 3 sin (x) = 0

3 sin^{2} (x) - 8 sin (x) - 3 = 0

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