What is the general solution for -1-3cos^2(x)+2cos(x)sqrt(3)=0 ?

The general solution for -1-3cos^2(x)+2cos(x)sqrt(3)=0 is x=0.95531…+2pin,x=2pi-0.95531…+2pin

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Popular >

-1-3cos^2(x)+2cos(x)sqrt(3)=0

Solution

- 1 - 3 cos^{2} (x) + 2 cos (x) 3 = 0

Solution

x = 0.95531 \dots + 2 πn, x = 2 π - 0.95531 \dots + 2 πn

Degrees

x = 54.73561 \dots^{\circ} + 36 0^{\circ} n, x = 305.26438 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 1 - 3 cos^{2} (x) + 2 cos (x) 3 = 0

Solve by substitution

- 1 - 3 cos^{2} (x) + 2 cos (x) 3 = 0

Let:

cos (x) = u

- 1 - 3 u^{2} + 2 u 3 = 0

- 1 - 3 u^{2} + 2 u 3 = 0 : u = \frac{3}{3}

- 1 - 3 u^{2} + 2 u 3 = 0

Write in the standard form

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

- 3 u^{2} + 23 u - 1 = 0

Solve with the quadratic formula

- 3 u^{2} + 23 u - 1 = 0

Quadratic Equation Formula:

For

a = - 3, b = 23, c = - 1

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

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

(23)^{2} - 4 (- 3) (- 1) = 0

(23)^{2} - 4 (- 3) (- 1)

Apply rule

- (- a) = a

= (23)^{2} - 4 \cdot 3 \cdot 1

(23)^{2} = 2^{2} \cdot 3

(23)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (3)^{2}

(3)^{2} : 3

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

= 2^{2} \cdot 3

4 \cdot 3 \cdot 1 = 12

4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 12

= 2^{2} \cdot 3 - 12

2^{2} \cdot 3 = 12

2^{2} \cdot 3

2^{2} = 4

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 12 - 12

Subtract the numbers:

12 - 12 = 0

= 0

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

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 3 = 6

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

Apply the fraction rule:

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

= \frac{2 3}{6}

Cancel the common factor:

2

= \frac{3}{3}

u = \frac{3}{3}

The solution to the quadratic equation is:

u = \frac{3}{3}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{3}{3} : x = arccos (\frac{3}{3}) + 2 πn, x = 2 π - arccos (\frac{3}{3}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3}{3}) + 2 πn, x = 2 π - arccos (\frac{3}{3}) + 2 πn

x = arccos (\frac{3}{3}) + 2 πn, x = 2 π - arccos (\frac{3}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{3}{3}) + 2 πn, x = 2 π - arccos (\frac{3}{3}) + 2 πn

Show solutions in decimal form

x = 0.95531 \dots + 2 πn, x = 2 π - 0.95531 \dots + 2 πn

Popular Examples

tan(3x)+cos(6x)=1

tan (3 x) + cos (6 x) = 1

sec(θ)+tan(θ)=2

sec (θ) + tan (θ) = 2

sin^2(θ)=2cos(θ)+2,0<= θ<= 2pi

sin^{2} (θ) = 2 cos (θ) + 2, 0 \leq θ \leq 2 π

sin(x)=-0.3

sin (x) = - 0.3

-sin(x)+cos(x)=0,-pi<= x<= pi

- sin (x) + cos (x) = 0, - π \leq x \leq π

Frequently Asked Questions (FAQ)

What is the general solution for -1-3cos^2(x)+2cos(x)sqrt(3)=0 ?
The general solution for -1-3cos^2(x)+2cos(x)sqrt(3)=0 is x=0.95531…+2pin,x=2pi-0.95531…+2pin