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

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

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4sin(θ)=-4cos(θ)

4 sin (θ) = - 4 cos (θ)

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

4 sin (θ) = - 4 cos (θ)

Rewrite using trig identities

4 sin (θ) = - 4 cos (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{4 sin ( θ )}{cos ( θ )} = \frac{- 4 cos ( θ )}{cos ( θ )}

Simplify

\frac{4 sin ( θ )}{cos ( θ )} = - 4

Use the basic trigonometric identity:

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

4 tan (θ) = - 4

4 tan (θ) = - 4

Divide both sides by

4

4 tan (θ) = - 4

Divide both sides by

4

\frac{4 tan ( θ )}{4} = \frac{- 4}{4}

Simplify

tan (θ) = - 1

tan (θ) = - 1

General solutions for

tan (θ) = - 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}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

- 3 sin (a) - 2 = sin (a) - 2

tan (A) = 1

3 sec^{2} (x) + 2 tan (x) - 23 = 0

sin (x) (1 - 2 cos (x)) = 0

4 cos^{2} (x) + 3 cos (x) = 1

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