What is the general solution for 2cos^2(a)-7cos(a)+3=0 ?

The general solution for 2cos^2(a)-7cos(a)+3=0 is a= pi/3+2pin,a=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(a)-7cos(a)+3=0

Solution

2 cos^{2} (a) - 7 cos (a) + 3 = 0

Solution

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

Degrees

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

Solution steps

2 cos^{2} (a) - 7 cos (a) + 3 = 0

Solve by substitution

2 cos^{2} (a) - 7 cos (a) + 3 = 0

Let:

cos (a) = u

2 u^{2} - 7 u + 3 = 0

2 u^{2} - 7 u + 3 = 0 : u = 3, u = \frac{1}{2}

2 u^{2} - 7 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

(- 7)^{2} - 4 \cdot 2 \cdot 3

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

u = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

\frac{- ( - 7 ) + 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 2}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 7 ) - 5}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 7 ) - 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 2}

Subtract the numbers:

7 - 5 = 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 = 3, u = \frac{1}{2}

Substitute back

u = cos (a)

cos (a) = 3, cos (a) = \frac{1}{2}

cos (a) = 3, cos (a) = \frac{1}{2}

cos (a) = 3 :

No Solution

cos (a) = 3

- 1 \leq cos (x) \leq 1

No Solution

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

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

General solutions for

cos (a) = \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}

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

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

Combine all the solutions

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

Popular Examples

solvefor y,x=2cos(2y)solve for

y, x = 2 cos (2 y)

-3-6cos(x)=0

- 3 - 6 cos (x) = 0

tan(θ)= 26/53

tan (θ) = \frac{26}{53}

solvefor n,y=-sin(2(pi/4+pin))2 solve for

n, y = - sin (2 (\frac{π}{4} + πn)) 2

tan(a)= 7/24

tan (a) = \frac{7}{24}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(a)-7cos(a)+3=0 ?
The general solution for 2cos^2(a)-7cos(a)+3=0 is a= pi/3+2pin,a=(5pi)/3+2pin