What is the general solution for 4cos(2θ)=4sin(2θ) ?

The general solution for 4cos(2θ)=4sin(2θ) is θ= pi/8+(pin)/2

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4cos(2θ)=4sin(2θ)

4 cos (2 θ) = 4 sin (2 θ)

θ = \frac{π}{8} + \frac{πn}{2}

Degrees

θ = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

4 cos (2 θ) = 4 sin (2 θ)

Rewrite using trig identities

4 cos (2 θ) = 4 sin (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

\frac{4 cos ( 2 θ )}{cos ( 2 θ )} = \frac{4 sin ( 2 θ )}{cos ( 2 θ )}

Simplify

4 = \frac{4 sin ( 2 θ )}{cos ( 2 θ )}

Use the basic trigonometric identity:

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

4 = 4 tan (2 θ)

Switch sides

4 tan (2 θ) = 4

4 tan (2 θ) = 4

Divide both sides by

4

4 tan (2 θ) = 4

Divide both sides by

4

\frac{4 tan ( 2 θ )}{4} = \frac{4}{4}

Simplify

tan (2 θ) = 1

tan (2 θ) = 1

General solutions for

tan (2 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}

2 θ = \frac{π}{4} + πn

2 θ = \frac{π}{4} + πn

Solve

2 θ = \frac{π}{4} + πn : θ = \frac{π}{8} + \frac{πn}{2}

2 θ = \frac{π}{4} + πn

Divide both sides by

2

2 θ = \frac{π}{4} + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

tan (x) = - 3, 0 \leq x \leq 2 π

5 tan^{2} (A) - 7 = 8

tan (x) = \frac{28}{16}

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

9 tan (2 x) - 18 cos (x) = 0

What is the general solution for 4cos(2θ)=4sin(2θ) ?
The general solution for 4cos(2θ)=4sin(2θ) is θ= pi/8+(pin)/2