What is the general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) ?

The general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) is θ=(5pi)/6+pin,θ=pin

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4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ)

Solution

4 tan^{2} (θ) = - 3 tan (θ) + tan^{2} (θ)

Solution

θ = \frac{5 π}{6} + πn, θ = πn

Degrees

θ = 15 0^{\circ} + 18 0^{\circ} n, θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (θ) = - 3 tan (θ) + tan^{2} (θ)

Solve by substitution

4 tan^{2} (θ) = - 3 tan (θ) + tan^{2} (θ)

Let:

tan (θ) = u

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

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

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

Switch sides

- 3 u + u^{2} = 4 u^{2}

Move

4 u^{2}

to the left side

- 3 u + u^{2} = 4 u^{2}

Subtract

4 u^{2}

from both sides

- 3 u + u^{2} - 4 u^{2} = 4 u^{2} - 4 u^{2}

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 3)^{2} = 3

(- 3)^{2}

Apply exponent rule:

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

n

is even

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

= (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 3 \cdot 0 = 0

4 \cdot 3 \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 ( - 3 )}

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add similar elements:

3 + 3 = 23

= \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}{2 ( - 3 )} : 0

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add similar elements:

3 - 3 = 0

= \frac{0}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{- 6}

Apply the fraction rule:

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

= - \frac{0}{6}

Apply rule

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

= - 0

= 0

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

tan (θ) = - \frac{3}{3}, tan (θ) = 0

tan (θ) = - \frac{3}{3}, tan (θ) = 0

tan (θ) = - \frac{3}{3} : θ = \frac{5 π}{6} + πn

tan (θ) = - \frac{3}{3}

General solutions for

tan (θ) = - \frac{3}{3}

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

Combine all the solutions

θ = \frac{5 π}{6} + πn, θ = πn

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3cos(2x)=-3

3 cos (2 x) = - 3

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4 cos^{2} (θ) + cos (2 θ) - 7 cos (θ) = - 2

cos(θ)=(4sqrt(2))/7

cos (θ) = \frac{4 2}{7}

sec^2(θ)+2sec(θ)=8

sec^{2} (θ) + 2 sec (θ) = 8

Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) ?
The general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) is θ=(5pi)/6+pin,θ=pin