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 = 231.33171 \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{1 + 17}{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

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 2

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 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

= \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{1 + 17}{4}

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

Remove parentheses:

(- 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 = - (1 + 17)

= \frac{1 + 17}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

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

Popular Examples

2cos^2(x)+sin(x)=5

2 cos^{2} (x) + sin (x) = 5

solvefor x,Y=0.5sin(3.07x-2.4t+0.59)solve for

x, Y = 0.5 sin (3.07 x - 2.4 t + 0.59)

((1+tan(x)))/(sec(x))=2,x

\frac{( 1 + tan ( x ) )}{sec ( x )} = 2, x

sin(x-30)cos(x-30)=(sqrt(3))/4

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

cos(θ)=-7/15 ,cos(θ/2),180<θ<270

cos (θ) = - \frac{7}{15}, cos (\frac{θ}{2}), 18 0^{\circ} < θ < 27 0^{\circ}

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