What is the general solution for 2cos^2(t)+7cos(t)+5=0 ?

The general solution for 2cos^2(t)+7cos(t)+5=0 is t=pi+2pin

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Popular >

2cos^2(t)+7cos(t)+5=0

Solution

2 cos^{2} (t) + 7 cos (t) + 5 = 0

Solution

t = π + 2 πn

Degrees

t = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (t) + 7 cos (t) + 5 = 0

Solve by substitution

2 cos^{2} (t) + 7 cos (t) + 5 = 0

Let:

cos (t) = u

2 u^{2} + 7 u + 5 = 0

2 u^{2} + 7 u + 5 = 0 : u = - 1, u = - \frac{5}{2}

2 u^{2} + 7 u + 5 = 0

Solve with the quadratic formula

2 u^{2} + 7 u + 5 = 0

Quadratic Equation Formula:

For

a = 2, b = 7, c = 5

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

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

7^{2} - 4 \cdot 2 \cdot 5

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 7^{2} - 40

7^{2} = 49

= 49 - 40

Subtract the numbers:

49 - 40 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 7 \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 7 + 3}{2 \cdot 2}, u_{2} = \frac{- 7 - 3}{2 \cdot 2}

u = \frac{- 7 + 3}{2 \cdot 2} : - 1

\frac{- 7 + 3}{2 \cdot 2}

Add/Subtract the numbers:

- 7 + 3 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

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

\frac{- 7 - 3}{2 \cdot 2}

Subtract the numbers:

- 7 - 3 = - 10

= \frac{- 10}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 10}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{10}{4}

Cancel the common factor:

2

= - \frac{5}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (t)

cos (t) = - 1, cos (t) = - \frac{5}{2}

cos (t) = - 1, cos (t) = - \frac{5}{2}

cos (t) = - 1 : t = π + 2 πn

cos (t) = - 1

General solutions for

cos (t) = - 1

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}

t = π + 2 πn

t = π + 2 πn

cos (t) = - \frac{5}{2} :

No Solution

cos (t) = - \frac{5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

t = π + 2 πn

Popular Examples

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cos (x) = \frac{35}{45}

pi/6 =cos(2t)

\frac{π}{6} = cos (2 t)

solvefor x,sqrt(3)tan(2x)=1 solve for

x, 3 tan (2 x) = 1

solvefor y,x=5*sin(sqrt(y))solve for

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12cos^2(x)-cos(x)-1=0

12 cos^{2} (x) - cos (x) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(t)+7cos(t)+5=0 ?
The general solution for 2cos^2(t)+7cos(t)+5=0 is t=pi+2pin