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

The general solution for 2sin^2(x)+cos^2(x)+cos(x)-1=0 is x=2.23703…+2pin,x=-2.23703…+2pin

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Popular >

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

Solution

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

Solution

x = 2.23703 \dots + 2 πn, x = - 2.23703 \dots + 2 πn

Degrees

x = 128.17270 \dots^{\circ} + 36 0^{\circ} n, x = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Simplify

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

1 + u - u^{2} = 0

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

1 + u - u^{2} = 0

Write in the standard form

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

- u^{2} + u + 1 = 0

Solve with the quadratic formula

- u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 1

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{- 1 + 5}{2} : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = - \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (x) = - \frac{- 1 + 5}{2}

General solutions for

cos (x) = - \frac{- 1 + 5}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{1 + 5}{2} :

No Solution

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = 2.23703 \dots + 2 πn, x = - 2.23703 \dots + 2 πn

Popular Examples

144^2=43.27^2+72^2-2*43.27*72*cos(x)

14 4^{2} = 43.2 7^{2} + 7 2^{2} - 2 \cdot 43.27 \cdot 72 \cdot cos (x)

sin(8x+48)=cos(-4x+26)

sin (8 x + 48) = cos (- 4 x + 26)

3cos(x)+sec(145)=1

3 cos (x) + sec (14 5^{\circ}) = 1

sec^2(x)+3tan^2(x)=13

sec^{2} (x) + 3 tan^{2} (x) = 13

1609.3=(1.133^2)/((9.8)tan(θ))

1609.3 = \frac{1.13 3 ^{2}}{( 9.8 ) tan ( θ )}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)+cos^2(x)+cos(x)-1=0 ?
The general solution for 2sin^2(x)+cos^2(x)+cos(x)-1=0 is x=2.23703…+2pin,x=-2.23703…+2pin