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

The general solution for 2cos^2(x)-sin(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Degrees

x = 51.33171 \dots^{\circ} + 36 0^{\circ} n, x = 128.66828 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- u + (1 - u^{2}) \cdot 2 = 0

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

- u + (1 - u^{2}) \cdot 2 = 0

Expand

- u + (1 - u^{2}) \cdot 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) \cdot 2

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

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

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

Apply rule

- (- a) = a

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

(- 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 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

Apply the fraction rule:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

Apply the fraction rule:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4} :

No Solution

sin (x) = - \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{17 - 1}{4} : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

sin (x) = \frac{17 - 1}{4}

Apply trig inverse properties

sin (x) = \frac{17 - 1}{4}

General solutions for

sin (x) = \frac{17 - 1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Graph

Popular Examples

4+4sin(x)=8cos(x)

4 + 4 sin (x) = 8 cos (x)

-3tan^2(θ)-4tan(θ)=-4

- 3 tan^{2} (θ) - 4 tan (θ) = - 4

(tan(x)+1)(sqrt(3)cot(x)+1)=0

(tan (x) + 1) (3 cot (x) + 1) = 0

4cos^2(x)-2=1

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

solvefor θ,x=rcos(θ)solve for

θ, x = r cos (θ)

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-sin(x)=0 ?
The general solution for 2cos^2(x)-sin(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin