What is the general solution for 7cos^2(x)+14cos(x)+7=0 ?

The general solution for 7cos^2(x)+14cos(x)+7=0 is x=pi+2pin

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7cos^2(x)+14cos(x)+7=0

7 cos^{2} (x) + 14 cos (x) + 7 = 0

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (x) + 14 cos (x) + 7 = 0

Solve by substitution

7 cos^{2} (x) + 14 cos (x) + 7 = 0

Let:

cos (x) = u

7 u^{2} + 14 u + 7 = 0

7 u^{2} + 14 u + 7 = 0 : u = - 1

7 u^{2} + 14 u + 7 = 0

Solve with the quadratic formula

7 u^{2} + 14 u + 7 = 0

Quadratic Equation Formula:

For

a = 7, b = 14, c = 7

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 7 \cdot 7}{2 \cdot 7}

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 7 \cdot 7}{2 \cdot 7}

1 4^{2} - 4 \cdot 7 \cdot 7 = 0

1 4^{2} - 4 \cdot 7 \cdot 7

Multiply the numbers:

4 \cdot 7 \cdot 7 = 196

= 1 4^{2} - 196

1 4^{2} = 196

= 196 - 196

Subtract the numbers:

196 - 196 = 0

= 0

u_{1, 2} = \frac{- 14 \pm 0}{2 \cdot 7}

u = \frac{- 14}{2 \cdot 7}

\frac{- 14}{2 \cdot 7} = - 1

\frac{- 14}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 14}{14}

Apply the fraction rule:

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

= - \frac{14}{14}

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 : x = π + 2 πn

cos (x) = - 1

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

Combine all the solutions

x = π + 2 πn

θ, 2 sin (θ) = 2 cos (2 θ)

sec (θ) + 4 = 4 sec (θ)

sin (2 x) - 4 \cdot cos (x) = 0

5 tan (θ) + 3 = 0

sin (θ) = - \frac{4}{7}, π < θ < \frac{3 π}{2}, cos (2 θ)

What is the general solution for 7cos^2(x)+14cos(x)+7=0 ?
The general solution for 7cos^2(x)+14cos(x)+7=0 is x=pi+2pin