What is the general solution for 2sin(x)=5cos(x),0<= x<360 ?

The general solution for 2sin(x)=5cos(x),0<= x<360 is x=1.19028…,x=1.19028…+180

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2sin(x)=5cos(x),0<= x<360

2 sin (x) = 5 cos (x), 0^{\circ} \leq x < 36 0^{\circ}

x = 1.19028 \dots, x = 1.19028 \dots + 18 0^{\circ}

Radians

x = 1.19028 \dots, x = 1.19028 \dots + π

Solution steps

2 sin (x) = 5 cos (x), 0^{\circ} \leq x < 36 0^{\circ}

Rewrite using trig identities

2 sin (x) = 5 cos (x)

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

2 tan (x) = 5

2 tan (x) = 5

Divide both sides by

2

2 tan (x) = 5

Divide both sides by

2

\frac{2 tan ( x )}{2} = \frac{5}{2}

Simplify

tan (x) = \frac{5}{2}

tan (x) = \frac{5}{2}

Apply trig inverse properties

tan (x) = \frac{5}{2}

General solutions for

tan (x) = \frac{5}{2}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

x = arctan (\frac{5}{2}) + 18 0^{\circ} n

x = arctan (\frac{5}{2}) + 18 0^{\circ} n

Solutions for the range

0 \leq x < 36 0^{\circ}

x = arctan (\frac{5}{2}), x = arctan (\frac{5}{2}) + 18 0^{\circ}

Show solutions in decimal form

x = 1.19028 \dots, x = 1.19028 \dots + 18 0^{\circ}

cos (x) = \frac{1 - 2}{26}

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

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

3 cot (θ) = 3

sec (x) = 1, - 2 π \leq x \leq 2 π

What is the general solution for 2sin(x)=5cos(x),0<= x<360 ?
The general solution for 2sin(x)=5cos(x),0<= x<360 is x=1.19028…,x=1.19028…+180