What is the general solution for 5tan^2(x)-9sec(x)+2=0 ?

The general solution for 5tan^2(x)-9sec(x)+2=0 is x=1.07121…+2pin,x=2pi-1.07121…+2pin

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Popular >

5tan^2(x)-9sec(x)+2=0

Solution

5 tan^{2} (x) - 9 sec (x) + 2 = 0

Solution

x = 1.07121 \dots + 2 πn, x = 2 π - 1.07121 \dots + 2 πn

Degrees

x = 61.37616 \dots^{\circ} + 36 0^{\circ} n, x = 298.62383 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 tan^{2} (x) - 9 sec (x) + 2 = 0

Rewrite using trig identities

2 + 5 tan^{2} (x) - 9 sec (x)

Use the Pythagorean identity:

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

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

= 2 + 5 (sec^{2} (x) - 1) - 9 sec (x)

Simplify

2 + 5 (sec^{2} (x) - 1) - 9 sec (x) : 5 sec^{2} (x) - 9 sec (x) - 3

2 + 5 (sec^{2} (x) - 1) - 9 sec (x)

Expand

5 (sec^{2} (x) - 1) : 5 sec^{2} (x) - 5

5 (sec^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = sec^{2} (x), c = 1

= 5 sec^{2} (x) - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 sec^{2} (x) - 5

= 2 + 5 sec^{2} (x) - 5 - 9 sec (x)

Simplify

2 + 5 sec^{2} (x) - 5 - 9 sec (x) : 5 sec^{2} (x) - 9 sec (x) - 3

2 + 5 sec^{2} (x) - 5 - 9 sec (x)

Group like terms

= 5 sec^{2} (x) - 9 sec (x) + 2 - 5

Add/Subtract the numbers:

2 - 5 = - 3

= 5 sec^{2} (x) - 9 sec (x) - 3

= 5 sec^{2} (x) - 9 sec (x) - 3

= 5 sec^{2} (x) - 9 sec (x) - 3

- 3 + 5 sec^{2} (x) - 9 sec (x) = 0

Solve by substitution

- 3 + 5 sec^{2} (x) - 9 sec (x) = 0

Let:

sec (x) = u

- 3 + 5 u^{2} - 9 u = 0

- 3 + 5 u^{2} - 9 u = 0 : u = \frac{9 + 141}{10}, u = \frac{9 - 141}{10}

- 3 + 5 u^{2} - 9 u = 0

Write in the standard form

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

5 u^{2} - 9 u - 3 = 0

Solve with the quadratic formula

5 u^{2} - 9 u - 3 = 0

Quadratic Equation Formula:

For

a = 5, b = - 9, c = - 3

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 9)^{2} - 4 \cdot 5 (- 3) = 141

(- 9)^{2} - 4 \cdot 5 (- 3)

Apply rule

- (- a) = a

= (- 9)^{2} + 4 \cdot 5 \cdot 3

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 9^{2} + 60

9^{2} = 81

= 81 + 60

Add the numbers:

81 + 60 = 141

= 141

u_{1, 2} = \frac{- ( - 9 ) \pm 141}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 141}{2 \cdot 5}, u_{2} = \frac{- ( - 9 ) - 141}{2 \cdot 5}

u = \frac{- ( - 9 ) + 141}{2 \cdot 5} : \frac{9 + 141}{10}

\frac{- ( - 9 ) + 141}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{9 + 141}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{9 + 141}{10}

u = \frac{- ( - 9 ) - 141}{2 \cdot 5} : \frac{9 - 141}{10}

\frac{- ( - 9 ) - 141}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{9 - 141}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{9 - 141}{10}

The solutions to the quadratic equation are:

u = \frac{9 + 141}{10}, u = \frac{9 - 141}{10}

Substitute back

u = sec (x)

sec (x) = \frac{9 + 141}{10}, sec (x) = \frac{9 - 141}{10}

sec (x) = \frac{9 + 141}{10}, sec (x) = \frac{9 - 141}{10}

sec (x) = \frac{9 + 141}{10} : x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

sec (x) = \frac{9 + 141}{10}

Apply trig inverse properties

sec (x) = \frac{9 + 141}{10}

General solutions for

sec (x) = \frac{9 + 141}{10}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

sec (x) = \frac{9 - 141}{10} :

No Solution

sec (x) = \frac{9 - 141}{10}

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

No Solution

Combine all the solutions

x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

Show solutions in decimal form

x = 1.07121 \dots + 2 πn, x = 2 π - 1.07121 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5tan^2(x)-9sec(x)+2=0 ?
The general solution for 5tan^2(x)-9sec(x)+2=0 is x=1.07121…+2pin,x=2pi-1.07121…+2pin