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=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

t = 2 πn

Degrees

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

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

Apply exponent rule:

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

n

is even

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

= 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:

n a^{n} = a

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} : \frac{5}{2}

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

7 - 3 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \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 = 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

Combine all the solutions

t = 2 πn

Graph

Popular Examples

(cos(x))=0

(cos (x)) = 0

arccos(x)= 3/8

arccos (x) = \frac{3}{8}

2sin^2(θ)-3sin(θ)=0

2 sin^{2} (θ) - 3 sin (θ) = 0

(cos(x))/(1+sin(x))+tan(x)=cos(x)

\frac{cos ( x )}{1 + sin ( x )} + tan (x) = cos (x)

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

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

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=2pin