What is the general solution for 4cos(x)=-sin^2(x)+1,0<= x<= 2pi ?

Q: What is the general solution for 4cos(x)=-sin^2(x)+1,0<= x<= 2pi ?

The general solution for 4cos(x)=-sin^2(x)+1,0<= x<= 2pi is x= pi/2 ,x=(3pi)/2

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4cos(x)=-sin^2(x)+1,0<= x<= 2pi

Solution

4 cos (x) = - sin^{2} (x) + 1, 0 \leq x \leq 2 π

Solution

x = \frac{π}{2}, x = \frac{3 π}{2}

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}

Solution steps

4 cos (x) = - sin^{2} (x) + 1, 0 \leq x \leq 2 π

Subtract

- sin^{2} (x) + 1

from both sides

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

Rewrite using trig identities

- 1 + sin^{2} (x) + 4 cos (x)

Use the Pythagorean identity:

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

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

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

- cos^{2} (x) + 4 cos (x) = 0

Solve by substitution

- cos^{2} (x) + 4 cos (x) = 0

Let:

cos (x) = u

- u^{2} + 4 u = 0

- u^{2} + 4 u = 0 : u = 0, u = 4

- u^{2} + 4 u = 0

Solve with the quadratic formula

- u^{2} + 4 u = 0

Quadratic Equation Formula:

For

a = - 1, b = 4, c = 0

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

4^{2} - 4 (- 1) \cdot 0 = 4

4^{2} - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} + 0

4^{2} + 0 = 4^{2}

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

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

Separate the solutions

u_{1} = \frac{- 4 + 4}{2 ( - 1 )}, u_{2} = \frac{- 4 - 4}{2 ( - 1 )}

u = \frac{- 4 + 4}{2 ( - 1 )} : 0

\frac{- 4 + 4}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 + 4}{- 2 \cdot 1}

Add/Subtract the numbers:

- 4 + 4 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

\frac{- 4 - 4}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 - 4}{- 2 \cdot 1}

Subtract the numbers:

- 4 - 4 = - 8

= \frac{- 8}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8}{- 2}

Apply the fraction rule:

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

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

The solutions to the quadratic equation are:

u = 0, u = 4

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = 4

cos (x) = 0, cos (x) = 4

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x \leq 2 π

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 4, 0 \leq x \leq 2 π :

No Solution

cos (x) = 4, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

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0=arcsin(x)

0 = arcsin (x)

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(x)=-sin^2(x)+1,0<= x<= 2pi ?
The general solution for 4cos(x)=-sin^2(x)+1,0<= x<= 2pi is x= pi/2 ,x=(3pi)/2