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=0,x=2pi

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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 = 0, x = 2 π

Degrees

x = 0^{\circ}, x = 36 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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 6)^{2} = 6^{2}

= 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}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 3 = 6

= \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 = 0, x = 2 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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

Combine all the solutions

x = 0, x = 2 π

Popular Examples

csc(θ)+2=0,0<= θ<= 2pi

csc (θ) + 2 = 0, 0 \leq θ \leq 2 π

sin^2(x)+5sin(x)-3=0

sin^{2} (x) + 5 sin (x) - 3 = 0

2-3sin(θ+(4pi)/3)=2

2 - 3 sin (θ + \frac{4 π}{3}) = 2

tan(θ)= 2/6

tan (θ) = \frac{2}{6}

4+3cos(x)=0

4 + 3 cos (x) = 0

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=0,x=2pi