What is the general solution for 5sin(x)cot(x)=1 ?

The general solution for 5sin(x)cot(x)=1 is x=1.36943…+2pin,x=2pi-1.36943…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5sin(x)cot(x)=1

5 sin (x) cot (x) = 1

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn

Degrees

x = 78.46304 \dots^{\circ} + 36 0^{\circ} n, x = 281.53695 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (x) cot (x) = 1

Subtract

1

from both sides

5 sin (x) cot (x) - 1 = 0

Rewrite using trig identities

- 1 + 5 cot (x) sin (x)

Use the basic trigonometric identity:

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

= - 1 + 5 \cdot \frac{cos ( x )}{sin ( x )} sin (x)

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

5 \cdot \frac{cos ( x )}{sin ( x )} sin (x)

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= cos (x) \cdot 5

= - 1 + 5 cos (x)

- 1 + 5 cos (x) = 0

Move

1

to the right side

- 1 + 5 cos (x) = 0

Add

1

to both sides

- 1 + 5 cos (x) + 1 = 0 + 1

Simplify

5 cos (x) = 1

5 cos (x) = 1

Divide both sides by

5

5 cos (x) = 1

Divide both sides by

5

\frac{5 cos ( x )}{5} = \frac{1}{5}

Simplify

cos (x) = \frac{1}{5}

cos (x) = \frac{1}{5}

Apply trig inverse properties

cos (x) = \frac{1}{5}

General solutions for

cos (x) = \frac{1}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

Show solutions in decimal form

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn

7 cos (7 x) = 0

0.75 = sin (x)

7 cos^{2} (x) - 1 = 5 cos (x), 0 \leq x \leq 2 π

2 cos^{2} (θ) = 3 sin (θ) + 3

cos (π x^{2}) = 0

What is the general solution for 5sin(x)cot(x)=1 ?
The general solution for 5sin(x)cot(x)=1 is x=1.36943…+2pin,x=2pi-1.36943…+2pin