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

The general solution for 2sin^2(x)-cos(x)+1=0 is x=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(x)-cos(x)+1=0

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

1 - cos (x) + 2 (1 - cos^{2} (x))

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

= 2 \cdot 1 - 2 cos^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

1 - cos (x) + 2 - 2 cos^{2} (x)

Group like terms

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

Add the numbers:

1 + 2 = 3

= - 2 cos^{2} (x) - cos (x) + 3

= - 2 cos^{2} (x) - cos (x) + 3

= - 2 cos^{2} (x) - cos (x) + 3

3 - cos (x) - 2 cos^{2} (x) = 0

Solve by substitution

3 - cos (x) - 2 cos^{2} (x) = 0

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 3

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

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

(- 1)^{2} - 4 (- 2) \cdot 3 = 5

(- 1)^{2} - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 3

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 3 = 24

4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 5}{2 ( - 2 )} : - \frac{3}{2}

\frac{- ( - 1 ) + 5}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 5}{- 2 \cdot 2}

Add the numbers:

1 + 5 = 6

= \frac{6}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{- 4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

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

\frac{- ( - 1 ) - 5}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 5}{- 2 \cdot 2}

Subtract the numbers:

1 - 5 = - 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{3}{2}, u = 1

Substitute back

u = cos (x)

cos (x) = - \frac{3}{2}, cos (x) = 1

cos (x) = - \frac{3}{2}, cos (x) = 1

cos (x) = - \frac{3}{2} :

No Solution

cos (x) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = 2 πn

Popular Examples

sin(θ)=-0.4

sin (θ) = - 0.4

sin(θ)=-0.2

sin (θ) = - 0.2

6cos(2x)=5

6 cos (2 x) = 5

2cos^2(2x)=1

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

sin(z)= 1/2

sin (z) = \frac{1}{2}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-cos(x)+1=0 ?
The general solution for 2sin^2(x)-cos(x)+1=0 is x=2pin