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

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

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Popular Trigonometry >

2tan(x)=3csc(x)

Solution

2 tan (x) = 3 csc (x)

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 tan (x) = 3 csc (x)

Subtract

3 csc (x)

from both sides

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

Express with sin, cos

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

Simplify

2 \cdot \frac{sin ( x )}{cos ( x )} - 3 \cdot \frac{1}{sin ( x )} : \frac{2 sin ^{2} ( x ) - 3 cos ( x )}{cos ( x ) sin ( x )}

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

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

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

Multiply fractions:

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

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

3 \cdot \frac{1}{sin ( x )} = \frac{3}{sin ( x )}

3 \cdot \frac{1}{sin ( x )}

Multiply fractions:

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

= \frac{1 \cdot 3}{sin ( x )}

Multiply the numbers:

1 \cdot 3 = 3

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

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

Least Common Multiplier of

cos (x), sin (x) : cos (x) sin (x)

cos (x), sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

sin (x)

= cos (x) sin (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

cos (x) sin (x)

For

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

multiply the denominator and numerator by

sin (x)

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

For

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

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Add

3 cos (x)

to both sides

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

Square both sides

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

Subtract

(3 cos (x))^{2}

from both sides

4 sin^{4} (x) - 9 cos^{2} (x) = 0

Factor

4 sin^{4} (x) - 9 cos^{2} (x) : (2 sin^{2} (x) + 3 cos (x)) (2 sin^{2} (x) - 3 cos (x))

4 sin^{4} (x) - 9 cos^{2} (x)

Rewrite

4 sin^{4} (x) - 9 cos^{2} (x)

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

4 sin^{4} (x) - 9 cos^{2} (x)

Rewrite

4

2^{2}

= 2^{2} sin^{4} (x) - 9 cos^{2} (x)

Rewrite

9

3^{2}

= 2^{2} sin^{4} (x) - 3^{2} cos^{2} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

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

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

Solving each part separately

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

2 sin^{2} (x) + 3 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 2 (1 - cos^{2} (x)) + 3 cos (x)

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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:

n a^{n} = a

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 )} : - \frac{1}{2}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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

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{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

2 sin^{2} (x) - 3 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (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:

n a^{n} = a

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 = cos (x)

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

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

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

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

Combine all the solutions

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

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 tan (x) = 3 csc (x)

Remove the ones that don't agree with the equation.

Check the solution

\frac{2 π}{3} + 2 πn :

False

\frac{2 π}{3} + 2 πn

Plug in

n = 1

\frac{2 π}{3} + 2 π 1

For

2 tan (x) = 3 csc (x)

plug in

x = \frac{2 π}{3} + 2 π 1

2 tan (\frac{2 π}{3} + 2 π 1) = 3 csc (\frac{2 π}{3} + 2 π 1)

Refine

- 3.46410 \dots = 3.46410 \dots

\Rightarrow False

Check the solution

\frac{4 π}{3} + 2 πn :

False

\frac{4 π}{3} + 2 πn

Plug in

n = 1

\frac{4 π}{3} + 2 π 1

For

2 tan (x) = 3 csc (x)

plug in

x = \frac{4 π}{3} + 2 π 1

2 tan (\frac{4 π}{3} + 2 π 1) = 3 csc (\frac{4 π}{3} + 2 π 1)

Refine

3.46410 \dots = - 3.46410 \dots

\Rightarrow False

Check the solution

\frac{π}{3} + 2 πn :

True

\frac{π}{3} + 2 πn

Plug in

n = 1

\frac{π}{3} + 2 π 1

For

2 tan (x) = 3 csc (x)

plug in

x = \frac{π}{3} + 2 π 1

2 tan (\frac{π}{3} + 2 π 1) = 3 csc (\frac{π}{3} + 2 π 1)

Refine

3.46410 \dots = 3.46410 \dots

\Rightarrow True

Check the solution

\frac{5 π}{3} + 2 πn :

True

\frac{5 π}{3} + 2 πn

Plug in

n = 1

\frac{5 π}{3} + 2 π 1

For

2 tan (x) = 3 csc (x)

plug in

x = \frac{5 π}{3} + 2 π 1

2 tan (\frac{5 π}{3} + 2 π 1) = 3 csc (\frac{5 π}{3} + 2 π 1)

Refine

- 3.46410 \dots = - 3.46410 \dots

\Rightarrow True

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

Graph

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Frequently Asked Questions (FAQ)

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