What is the general solution for (cos(x))/(tan(x))= 3/2 ?

The general solution for (cos(x))/(tan(x))= 3/2 is x= pi/6+2pin,x=(5pi)/6+2pin

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(cos(x))/(tan(x))= 3/2

Solution

\frac{cos ( x )}{tan ( x )} = \frac{3}{2}

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{cos ( x )}{tan ( x )} = \frac{3}{2}

Subtract

\frac{3}{2}

from both sides

\frac{cos ( x )}{tan ( x )} - \frac{3}{2} = 0

Simplify

\frac{cos ( x )}{tan ( x )} - \frac{3}{2} : \frac{2 cos ( x ) - 3 tan ( x )}{2 tan ( x )}

\frac{cos ( x )}{tan ( x )} - \frac{3}{2}

Least Common Multiplier of

tan (x), 2 : 2 tan (x)

tan (x), 2

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

tan (x)

2

= 2 tan (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

2 tan (x)

For

\frac{cos ( x )}{tan ( x )} :

multiply the denominator and numerator by

2

\frac{cos ( x )}{tan ( x )} = \frac{cos ( x ) \cdot 2}{tan ( x ) \cdot 2}

For

\frac{3}{2} :

multiply the denominator and numerator by

tan (x)

\frac{3}{2} = \frac{3 tan ( x )}{2 tan ( x )}

= \frac{cos ( x ) \cdot 2}{tan ( x ) \cdot 2} - \frac{3 tan ( x )}{2 tan ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( x ) \cdot 2 - 3 tan ( x )}{2 tan ( x )}

\frac{2 cos ( x ) - 3 tan ( x )}{2 tan ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Express with sin, cos

2 cos (x) - 3 tan (x)

Use the basic trigonometric identity:

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

= 2 cos (x) - 3 \cdot \frac{sin ( x )}{cos ( x )}

Simplify

2 cos (x) - 3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{2 cos ^{2} ( x ) - 3 sin ( x )}{cos ( x )}

2 cos (x) - 3 \cdot \frac{sin ( x )}{cos ( x )}

Multiply

3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{3 sin ( x )}{cos ( x )}

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

Multiply fractions:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

= 2 cos (x) - \frac{3 sin ( x )}{cos ( x )}

Convert element to fraction:

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

= \frac{2 cos ( x ) cos ( x )}{cos ( x )} - \frac{sin ( x ) \cdot 3}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 cos ( x ) cos ( x ) - sin ( x ) \cdot 3}{cos ( x )}

2 cos (x) cos (x) - sin (x) \cdot 3 = 2 cos^{2} (x) - 3 sin (x)

2 cos (x) cos (x) - sin (x) \cdot 3

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

2 cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= 2 cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (x)

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

= \frac{2 cos ^{2} ( x ) - 3 sin ( x )}{cos ( x )}

= \frac{2 cos ^{2} ( x ) - 3 sin ( x )}{cos ( x )}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 2 (1 - sin^{2} (x)) - 3 sin (x)

(1 - sin^{2} (x)) \cdot 2 - 3 sin (x) = 0

Solve by substitution

(1 - sin^{2} (x)) \cdot 2 - 3 sin (x) = 0

Let:

sin (x) = u

(1 - u^{2}) \cdot 2 - 3 u = 0

(1 - u^{2}) \cdot 2 - 3 u = 0 : u = - 2, u = \frac{1}{2}

(1 - u^{2}) \cdot 2 - 3 u = 0

Expand

(1 - u^{2}) \cdot 2 - 3 u : 2 - 2 u^{2} - 3 u

(1 - u^{2}) \cdot 2 - 3 u

= 2 (1 - u^{2}) - 3 u

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= 2 - 2 u^{2} - 3 u

2 - 2 u^{2} - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} - 3 u + 2 = 0

Solve with the quadratic formula

- 2 u^{2} - 3 u + 2 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 3, c = 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

(- 3)^{2} - 4 (- 2) \cdot 2

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 2 \cdot 2

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 3^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 + 5}{- 2 \cdot 2}

Add the numbers:

3 + 5 = 8

= \frac{8}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 - 5}{- 2 \cdot 2}

Subtract the numbers:

3 - 5 = - 2

= \frac{- 2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - 2, u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = - 2, sin (x) = \frac{1}{2}

sin (x) = - 2, sin (x) = \frac{1}{2}

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

cos(2x)+5cos(x)=3

cos (2 x) + 5 cos (x) = 3

sin(θ)=(207.91*sin(31.76977))/(238.99)

sin (θ) = \frac{207.91 \cdot sin ( 31.7697 7 ^{\circ} )}{238.99}

15=-40sin(18(x-10))

15 = - 40 sin (18 (x - 10))

tan^2(x)-tan(x)-6=0

tan^{2} (x) - tan (x) - 6 = 0

1/(sec(2x)-1)-1/(sec(2x)+1)=6

\frac{1}{sec ( 2 x ) - 1} - \frac{1}{sec ( 2 x ) + 1} = 6

Frequently Asked Questions (FAQ)

What is the general solution for (cos(x))/(tan(x))= 3/2 ?
The general solution for (cos(x))/(tan(x))= 3/2 is x= pi/6+2pin,x=(5pi)/6+2pin