What is the general solution for 2csc(x)cot(x)= 4/3 ?

The general solution for 2csc(x)cot(x)= 4/3 is x= pi/3+2pin,x=(5pi)/3+2pin

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2csc(x)cot(x)= 4/3

Solution

2 csc (x) cot (x) = \frac{4}{3}

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 csc (x) cot (x) = \frac{4}{3}

Subtract

\frac{4}{3}

from both sides

2 csc (x) cot (x) - \frac{4}{3} = 0

Simplify

2 csc (x) cot (x) - \frac{4}{3} : \frac{6 csc ( x ) cot ( x ) - 4}{3}

2 csc (x) cot (x) - \frac{4}{3}

Convert element to fraction:

2 csc (x) cot (x) = \frac{2 csc ( x ) cot ( x ) 3}{3}

= \frac{2 csc ( x ) cot ( x ) \cdot 3}{3} - \frac{4}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 csc ( x ) cot ( x ) \cdot 3 - 4}{3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6 csc ( x ) cot ( x ) - 4}{3}

\frac{6 csc ( x ) cot ( x ) - 4}{3} = 0

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

6 csc (x) cot (x) - 4 = 0

Express with sin, cos

- 4 + 6 cot (x) csc (x)

Use the basic trigonometric identity:

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

= - 4 + 6 \cdot \frac{cos ( x )}{sin ( x )} csc (x)

Use the basic trigonometric identity:

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

= - 4 + 6 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )}

Simplify

- 4 + 6 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )} : \frac{- 4 sin ^{2} ( x ) + 6 cos ( x )}{sin ^{2} ( x )}

- 4 + 6 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )}

6 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )} = \frac{6 cos ( x )}{sin ^{2} ( x )}

6 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )}

Multiply fractions:

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

= \frac{cos ( x ) \cdot 1 \cdot 6}{sin ( x ) sin ( x )}

Multiply the numbers:

1 \cdot 6 = 6

= \frac{6 cos ( x )}{sin ( x ) sin ( x )}

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= \frac{6 cos ( x )}{sin ^{2} ( x )}

= - 4 + \frac{6 cos ( x )}{sin ^{2} ( x )}

Convert element to fraction:

4 = \frac{4 sin ^{2} ( x )}{sin ^{2} ( x )}

= - \frac{4 sin ^{2} ( x )}{sin ^{2} ( x )} + \frac{6 cos ( x )}{sin ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 4 sin ^{2} ( x ) + 6 cos ( x )}{sin ^{2} ( x )}

= \frac{- 4 sin ^{2} ( x ) + 6 cos ( x )}{sin ^{2} ( x )}

\frac{- 4 sin ^{2} ( x ) + 6 cos ( x )}{sin ^{2} ( x )} = 0

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

- 4 sin^{2} (x) + 6 cos (x) = 0

Rewrite using trig identities

- 4 sin^{2} (x) + 6 cos (x)

Use the Pythagorean identity:

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

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

= - 4 (1 - cos^{2} (x)) + 6 cos (x)

- (1 - cos^{2} (x)) \cdot 4 + 6 cos (x) = 0

Solve by substitution

- (1 - cos^{2} (x)) \cdot 4 + 6 cos (x) = 0

Let:

cos (x) = u

- (1 - u^{2}) \cdot 4 + 6 u = 0

- (1 - u^{2}) \cdot 4 + 6 u = 0 : u = \frac{1}{2}, u = - 2

- (1 - u^{2}) \cdot 4 + 6 u = 0

Expand

- (1 - u^{2}) \cdot 4 + 6 u : - 4 + 4 u^{2} + 6 u

- (1 - u^{2}) \cdot 4 + 6 u

= - 4 (1 - u^{2}) + 6 u

Expand

- 4 (1 - u^{2}) : - 4 + 4 u^{2}

- 4 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = u^{2}

= - 4 \cdot 1 - (- 4) u^{2}

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 u^{2}

= - 4 + 4 u^{2} + 6 u

- 4 + 4 u^{2} + 6 u = 0

Write in the standard form

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

4 u^{2} + 6 u - 4 = 0

Solve with the quadratic formula

4 u^{2} + 6 u - 4 = 0

Quadratic Equation Formula:

For

a = 4, b = 6, c = - 4

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

6^{2} - 4 \cdot 4 (- 4) = 10

6^{2} - 4 \cdot 4 (- 4)

Apply rule

- (- a) = a

= 6^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 6^{2} + 64

6^{2} = 36

= 36 + 64

Add the numbers:

36 + 64 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 6 \pm 10}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 6 + 10}{2 \cdot 4}, u_{2} = \frac{- 6 - 10}{2 \cdot 4}

u = \frac{- 6 + 10}{2 \cdot 4} : \frac{1}{2}

\frac{- 6 + 10}{2 \cdot 4}

Add/Subtract the numbers:

- 6 + 10 = 4

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 6 - 10}{2 \cdot 4} : - 2

\frac{- 6 - 10}{2 \cdot 4}

Subtract the numbers:

- 6 - 10 = - 16

= \frac{- 16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{8}

Apply the fraction rule:

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

= - \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

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

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

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cos (t) = \frac{24}{25}

sec^2(t)+2sec(t)=0

sec^{2} (t) + 2 sec (t) = 0

sin(270+x)-cos(180-x)=-sin(x)

sin (27 0^{\circ} + x) - cos (18 0^{\circ} - x) = - sin (x)

9cos(3x)=0

9 cos (3 x) = 0

16(l)cos(θ)=5sin(2θ)(2l)

16 (l) cos (θ) = 5 sin (2 θ) (2 l)

Frequently Asked Questions (FAQ)

What is the general solution for 2csc(x)cot(x)= 4/3 ?
The general solution for 2csc(x)cot(x)= 4/3 is x= pi/3+2pin,x=(5pi)/3+2pin