What is the general solution for 2sec^2(x)-5sec(x)+2=0,0<= x<= 360 ?

Q: What is the general solution for 2sec^2(x)-5sec(x)+2=0,0<= x<= 360 ?

The general solution for 2sec^2(x)-5sec(x)+2=0,0<= x<= 360 is x=60,x=300

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sec^2(x)-5sec(x)+2=0,0<= x<= 360

Solution

2 sec^{2} (x) - 5 sec (x) + 2 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 6 0^{\circ}, x = 30 0^{\circ}

Radians

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

2 sec^{2} (x) - 5 sec (x) + 2 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Solve by substitution

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

Let:

sec (x) = u

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

2 u^{2} - 5 u + 2 = 0 : u = 2, u = \frac{1}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 5, c = 2

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

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

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

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

Apply exponent rule:

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

n

is even

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

= 5^{2} - 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

5 - 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \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, 0 \leq x \leq 36 0^{\circ} : x = 6 0^{\circ}, x = 30 0^{\circ}

sec (x) = 2, 0 \leq x \leq 36 0^{\circ}

General solutions for

sec (x) = 2

sec (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

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

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

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 6 0^{\circ}, x = 30 0^{\circ}

sec (x) = \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} :

No Solution

sec (x) = \frac{1}{2}, 0 \leq x \leq 36 0^{\circ}

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

No Solution

Combine all the solutions

x = 6 0^{\circ}, x = 30 0^{\circ}

Popular Examples

2sin(5x)+sqrt(3)=0,0<= x<= 2pi

2 sin (5 x) + 3 = 0, 0 \leq x \leq 2 π

sin(2θ)=-24/25

sin (2 θ) = - \frac{24}{25}

sin(4x)+cos(4x)=1

sin (4 x) + cos (4 x) = 1

-4cos(x)=-sin^2(x)+4,0<= x<= 2pi

- 4 cos (x) = - sin^{2} (x) + 4, 0 \leq x \leq 2 π

0=cos^2(pi/(12)t)

0 = cos^{2} (\frac{π}{12} t)

Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(x)-5sec(x)+2=0,0<= x<= 360 ?
The general solution for 2sec^2(x)-5sec(x)+2=0,0<= x<= 360 is x=60,x=300