What is the general solution for 3tan^2(x)-8tan(x)+5=0 ?

The general solution for 3tan^2(x)-8tan(x)+5=0 is x=1.03037…+pin,x= pi/4+pin

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3tan^2(x)-8tan(x)+5=0

Solution

3 tan^{2} (x) - 8 tan (x) + 5 = 0

Solution

x = 1.03037 \dots + πn, x = \frac{π}{4} + πn

Degrees

x = 59.03624 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

3 tan^{2} (x) - 8 tan (x) + 5 = 0

Solve by substitution

3 tan^{2} (x) - 8 tan (x) + 5 = 0

Let:

tan (x) = u

3 u^{2} - 8 u + 5 = 0

3 u^{2} - 8 u + 5 = 0 : u = \frac{5}{3}, u = 1

3 u^{2} - 8 u + 5 = 0

Solve with the quadratic formula

3 u^{2} - 8 u + 5 = 0

Quadratic Equation Formula:

For

a = 3, b = - 8, c = 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 \cdot 5 = 2

(- 8)^{2} - 4 \cdot 3 \cdot 5

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 3 \cdot 5

Multiply the numbers:

4 \cdot 3 \cdot 5 = 60

= 8^{2} - 60

8^{2} = 64

= 64 - 60

Subtract the numbers:

64 - 60 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 8 ) \pm 2}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 2}{2 \cdot 3}, u_{2} = \frac{- ( - 8 ) - 2}{2 \cdot 3}

u = \frac{- ( - 8 ) + 2}{2 \cdot 3} : \frac{5}{3}

\frac{- ( - 8 ) + 2}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 + 2}{2 \cdot 3}

Add the numbers:

8 + 2 = 10

= \frac{10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{10}{6}

Cancel the common factor:

2

= \frac{5}{3}

u = \frac{- ( - 8 ) - 2}{2 \cdot 3} : 1

\frac{- ( - 8 ) - 2}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 - 2}{2 \cdot 3}

Subtract the numbers:

8 - 2 = 6

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{5}{3}, u = 1

Substitute back

u = tan (x)

tan (x) = \frac{5}{3}, tan (x) = 1

tan (x) = \frac{5}{3}, tan (x) = 1

tan (x) = \frac{5}{3} : x = arctan (\frac{5}{3}) + πn

tan (x) = \frac{5}{3}

Apply trig inverse properties

tan (x) = \frac{5}{3}

General solutions for

tan (x) = \frac{5}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{5}{3}) + πn

x = arctan (\frac{5}{3}) + πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = arctan (\frac{5}{3}) + πn, x = \frac{π}{4} + πn

Show solutions in decimal form

x = 1.03037 \dots + πn, x = \frac{π}{4} + πn

Popular Examples

sin(2θ)=-24/25 ,sin(θ)= 3/5

sin (2 θ) = - \frac{24}{25}, sin (θ) = \frac{3}{5}

100+60sin((7pi)/3 x)=110,0<= x<= 1

100 + 60 sin (\frac{7 π}{3} x) = 110, 0 \leq x \leq 1

sin(4x)= 1/(sqrt(2))

sin (4 x) = \frac{1}{2}

cot^2(x)+csc(x)=1,0<= x<2pi

cot^{2} (x) + csc (x) = 1, 0 \leq x < 2 π

81pi=10.5-9cos(pi/(30)t)

81 π = 10.5 - 9 cos (\frac{π}{30} t)

Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(x)-8tan(x)+5=0 ?
The general solution for 3tan^2(x)-8tan(x)+5=0 is x=1.03037…+pin,x= pi/4+pin