What is the general solution for 2tan^2(x)-3sec(x)=0 ?

The general solution for 2tan^2(x)-3sec(x)=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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2tan^2(x)-3sec(x)=0

Solution

2 tan^{2} (x) - 3 sec (x) = 0

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^{2} (x) - 3 sec (x) = 0

Rewrite using trig identities

2 tan^{2} (x) - 3 sec (x)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

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

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

Solve by substitution

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

Let:

sec (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 (- 1) + 2 u^{2}

Apply minus-plus rules

+ (- a) = - a

= - 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 \cdot 2 ( - 2 )}{2 \cdot 2}

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

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

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

Separate the solutions

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

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

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

Apply rule

- (- 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}

Divide the numbers:

\frac{8}{4} = 2

= 2

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

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

Apply rule

- (- 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 = sec (x)

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

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

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

sec (x) = 2

General solutions for

sec (x) = 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

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

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

No Solution

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

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(x)-3sec(x)=0 ?
The general solution for 2tan^2(x)-3sec(x)=0 is x= pi/3+2pin,x=(5pi)/3+2pin