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=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 = 2 πn

Degrees

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

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

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} : 1

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

Add/Subtract the numbers:

- 3 + 7 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Subtract the numbers:

- 3 - 7 = - 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 = 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 = 0 + 2 πn

t = 0 + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 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 = 2 πn

Popular Examples

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6 cos^{2} (x) + 7 sin (x) - 8 = 0

sqrt(3csc(x))=2

3 csc (x) = 2

4csc^2(x)-25=0

4 csc^{2} (x) - 25 = 0

sin(θ)=-4/5 ,(3pi)/2 <θ<2pi

sin (θ) = - \frac{4}{5}, \frac{3 π}{2} < θ < 2 π

4cos(x)+sin^2(x)=3cos^2(x)+2

4 cos (x) + sin^{2} (x) = 3 cos^{2} (x) + 2

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=2pin