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

The general solution for 2cos(x)-5sin(x)=0 is x=0.38050…+pin

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2cos(x)-5sin(x)=0

2 cos (x) - 5 sin (x) = 0

x = 0.38050 \dots + πn

Degrees

x = 21.80140 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (x) - 5 sin (x) = 0

Rewrite using trig identities

2 cos (x) - 5 sin (x) = 0

Divide both sides by

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

\frac{2 cos ( x ) - 5 sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

2 - \frac{5 sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

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

2 - 5 tan (x) = 0

2 - 5 tan (x) = 0

Move

2

to the right side

2 - 5 tan (x) = 0

Subtract

2

from both sides

2 - 5 tan (x) - 2 = 0 - 2

Simplify

- 5 tan (x) = - 2

- 5 tan (x) = - 2

Divide both sides by

- 5

- 5 tan (x) = - 2

Divide both sides by

- 5

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (\frac{2}{5}) + πn

x = arctan (\frac{2}{5}) + πn

Show solutions in decimal form

x = 0.38050 \dots + πn

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

sin (3 0^{\circ}) = cos (2 x)

tan (2 x) sin (x) - sin (x) - tan (2 x) + 1 = 0

3 sec (x) - cos (x) - 2 = 0

- 2 sin (t) = - 2

What is the general solution for 2cos(x)-5sin(x)=0 ?
The general solution for 2cos(x)-5sin(x)=0 is x=0.38050…+pin