What is the general solution for 2sin^2(θ)-5cos(θ)+1=0 ?

The general solution for 2sin^2(θ)-5cos(θ)+1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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2sin^2(θ)-5cos(θ)+1=0

Solution

2 sin^{2} (θ) - 5 cos (θ) + 1 = 0

Solution

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

Degrees

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

Solution steps

2 sin^{2} (θ) - 5 cos (θ) + 1 = 0

Rewrite using trig identities

1 + 2 sin^{2} (θ) - 5 cos (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + 2 (1 - cos^{2} (θ)) - 5 cos (θ)

Simplify

1 + 2 (1 - cos^{2} (θ)) - 5 cos (θ) : - 2 cos^{2} (θ) - 5 cos (θ) + 3

1 + 2 (1 - cos^{2} (θ)) - 5 cos (θ)

Expand

2 (1 - cos^{2} (θ)) : 2 - 2 cos^{2} (θ)

2 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (θ)

= 2 \cdot 1 - 2 cos^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (θ)

= 1 + 2 - 2 cos^{2} (θ) - 5 cos (θ)

Add the numbers:

1 + 2 = 3

= - 2 cos^{2} (θ) - 5 cos (θ) + 3

= - 2 cos^{2} (θ) - 5 cos (θ) + 3

3 - 2 cos^{2} (θ) - 5 cos (θ) = 0

Solve by substitution

3 - 2 cos^{2} (θ) - 5 cos (θ) = 0

Let:

cos (θ) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add the numbers:

5 + 7 = 12

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{- 4}

Apply the fraction rule:

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

= - \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= - 3

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Subtract the numbers:

5 - 7 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \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 (θ)

cos (θ) = - 3, cos (θ) = \frac{1}{2}

cos (θ) = - 3, cos (θ) = \frac{1}{2}

cos (θ) = - 3 :

No Solution

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

No Solution

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

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \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}

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

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

Combine all the solutions

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

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5 sin (x) - 4 = 2 cos (2 x)

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7=4sin(pi/2 (x-2))+5

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)-5cos(θ)+1=0 ?
The general solution for 2sin^2(θ)-5cos(θ)+1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin