What is the general solution for sqrt(3)cos(x)-2cos^2(x)=0,0<x<360 ?

Q: What is the general solution for sqrt(3)cos(x)-2cos^2(x)=0,0<x<360 ?

The general solution for sqrt(3)cos(x)-2cos^2(x)=0,0<x<360 is x=90,x=270,x=30,x=330

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sqrt(3)cos(x)-2cos^2(x)=0,0<x<360

Solution

3 cos (x) - 2 cos^{2} (x) = 0, 0^{\circ} < x < 36 0^{\circ}

Solution

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 3 0^{\circ}, x = 33 0^{\circ}

Radians

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{6}, x = \frac{11 π}{6}

Solution steps

3 cos (x) - 2 cos^{2} (x) = 0, 0^{\circ} < x < 36 0^{\circ}

Solve by substitution

3 cos (x) - 2 cos^{2} (x) = 0

Let:

cos (x) = u

3 u - 2 u^{2} = 0

3 u - 2 u^{2} = 0 : u = 0, u = \frac{3}{2}

3 u - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 3, c = 0

u_{1, 2} = \frac{- 3 \pm ( 3 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm ( 3 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(3)^{2} - 4 (- 2) \cdot 0 = 3

(3)^{2} - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= (3)^{2} + 4 \cdot 2 \cdot 0

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= 3

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 3 + 0

Add the numbers:

3 + 0 = 3

= 3

u_{1, 2} = \frac{- 3 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 3 + 3}{2 ( - 2 )}, u_{2} = \frac{- 3 - 3}{2 ( - 2 )}

u = \frac{- 3 + 3}{2 ( - 2 )} : 0

\frac{- 3 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 3}{- 2 \cdot 2}

Add similar elements:

- 3 + 3 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

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

= - \frac{0}{4}

Apply rule

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

= - 0

= 0

u = \frac{- 3 - 3}{2 ( - 2 )} : \frac{3}{2}

\frac{- 3 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Add similar elements:

- 3 - 3 = - 23

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 3}{- 4}

Apply the fraction rule:

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

= \frac{2 3}{4}

Cancel the common factor:

2

= \frac{3}{2}

The solutions to the quadratic equation are:

u = 0, u = \frac{3}{2}

Substitute back

u = cos (x)

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

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

cos (x) = 0, 0 < x < 36 0^{\circ} : x = 9 0^{\circ}, x = 27 0^{\circ}

cos (x) = 0, 0 < x < 36 0^{\circ}

General solutions for

cos (x) = 0

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 < x < 36 0^{\circ}

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

cos (x) = \frac{3}{2}, 0 < x < 36 0^{\circ} : x = 3 0^{\circ}, x = 33 0^{\circ}

cos (x) = \frac{3}{2}, 0 < x < 36 0^{\circ}

General solutions for

cos (x) = \frac{3}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 < x < 36 0^{\circ}

x = 3 0^{\circ}, x = 33 0^{\circ}

Combine all the solutions

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 3 0^{\circ}, x = 33 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)cos(x)-2cos^2(x)=0,0<x<360 ?
The general solution for sqrt(3)cos(x)-2cos^2(x)=0,0<x<360 is x=90,x=270,x=30,x=330