What is the general solution for 3cos^2(x)+6cos(x)=-3,0<= x<= 2pi ?

Q: What is the general solution for 3cos^2(x)+6cos(x)=-3,0<= x<= 2pi ?

The general solution for 3cos^2(x)+6cos(x)=-3,0<= x<= 2pi is x=pi

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3cos^2(x)+6cos(x)=-3,0<= x<= 2pi

Solution

3 cos^{2} (x) + 6 cos (x) = - 3, 0 \leq x \leq 2 π

Solution

x = π

Degrees

x = 18 0^{\circ}

Solution steps

3 cos^{2} (x) + 6 cos (x) = - 3, 0 \leq x \leq 2 π

Solve by substitution

3 cos^{2} (x) + 6 cos (x) = - 3

Let:

cos (x) = u

3 u^{2} + 6 u = - 3

3 u^{2} + 6 u = - 3 : u = - 1

3 u^{2} + 6 u = - 3

Move

3

to the left side

3 u^{2} + 6 u = - 3

Add

3

to both sides

3 u^{2} + 6 u + 3 = - 3 + 3

Simplify

3 u^{2} + 6 u + 3 = 0

3 u^{2} + 6 u + 3 = 0

Solve with the quadratic formula

3 u^{2} + 6 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = 6, c = 3

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

6^{2} - 4 \cdot 3 \cdot 3 = 0

6^{2} - 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 6^{2} - 36

6^{2} = 36

= 36 - 36

Subtract the numbers:

36 - 36 = 0

= 0

u_{1, 2} = \frac{- 6 \pm 0}{2 \cdot 3}

u = \frac{- 6}{2 \cdot 3}

\frac{- 6}{2 \cdot 3} = - 1

\frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

Combine all the solutions

x = π

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)+6cos(x)=-3,0<= x<= 2pi ?
The general solution for 3cos^2(x)+6cos(x)=-3,0<= x<= 2pi is x=pi