What is the general solution for 2cos^2(t)-3cos(t)-5=0 ?

The general solution for 2cos^2(t)-3cos(t)-5=0 is t=pi+2pin

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Popular >

2cos^2(t)-3cos(t)-5=0

Solution

2 cos^{2} (t) - 3 cos (t) - 5 = 0

Solution

t = π + 2 πn

Degrees

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

Solution steps

2 cos^{2} (t) - 3 cos (t) - 5 = 0

Solve by substitution

2 cos^{2} (t) - 3 cos (t) - 5 = 0

Let:

cos (t) = u

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

2 u^{2} - 3 u - 5 = 0 : u = \frac{5}{2}, u = - 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 3, c = - 5

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 3^{2} + 40

3^{2} = 9

= 9 + 40

Add the numbers:

9 + 40 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 7 = 10

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{10}{4}

Cancel the common factor:

2

= \frac{5}{2}

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 7 = - 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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (t)

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

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

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

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

Combine all the solutions

t = π + 2 πn

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Frequently Asked Questions (FAQ)

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