What is the general solution for sin(2x)=5cos(2x) ?

The general solution for sin(2x)=5cos(2x) is x=(1.37340…)/2+(pin)/2

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sin(2x)=5cos(2x)

sin (2 x) = 5 cos (2 x)

x = \frac{1.37340 \dots}{2} + \frac{πn}{2}

Degrees

x = 39.34503 \dots^{\circ} + 9 0^{\circ} n

Solution steps

sin (2 x) = 5 cos (2 x)

Rewrite using trig identities

sin (2 x) = 5 cos (2 x)

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

\frac{sin ( 2 x )}{cos ( 2 x )} = \frac{5 cos ( 2 x )}{cos ( 2 x )}

Simplify

\frac{sin ( 2 x )}{cos ( 2 x )} = 5

Use the basic trigonometric identity:

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

tan (2 x) = 5

tan (2 x) = 5

Apply trig inverse properties

tan (2 x) = 5

General solutions for

tan (2 x) = 5

tan (x) = a \Rightarrow x = arctan (a) + πn

2 x = arctan (5) + πn

2 x = arctan (5) + πn

Solve

2 x = arctan (5) + πn : x = \frac{arctan ( 5 )}{2} + \frac{πn}{2}

2 x = arctan (5) + πn

Divide both sides by

2

2 x = arctan (5) + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arctan ( 5 )}{2} + \frac{πn}{2}

Simplify

x = \frac{arctan ( 5 )}{2} + \frac{πn}{2}

x = \frac{arctan ( 5 )}{2} + \frac{πn}{2}

x = \frac{arctan ( 5 )}{2} + \frac{πn}{2}

Show solutions in decimal form

x = \frac{1.37340 \dots}{2} + \frac{πn}{2}

5 sec (2 x) + 2 = 0

2.1 sin (θ) = 1.33 sin (θ + 9 0^{\circ})

1 sin (4 5^{\circ}) = 1.33 sin (θ)

3 cot^{2} (x) = 9

1 2^{(2)} = 1 3^{(2)} + 5^{(2)} - 2 (13) (5) cos (B)

What is the general solution for sin(2x)=5cos(2x) ?
The general solution for sin(2x)=5cos(2x) is x=(1.37340…)/2+(pin)/2