What is the general solution for 4tan^2(x)-16tan(x)+7=0 ?

The general solution for 4tan^2(x)-16tan(x)+7=0 is x=1.29249…+pin,x=0.46364…+pin

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4tan^2(x)-16tan(x)+7=0

Solution

4 tan^{2} (x) - 16 tan (x) + 7 = 0

Solution

x = 1.29249 \dots + πn, x = 0.46364 \dots + πn

Degrees

x = 74.05460 \dots^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (x) - 16 tan (x) + 7 = 0

Solve by substitution

4 tan^{2} (x) - 16 tan (x) + 7 = 0

Let:

tan (x) = u

4 u^{2} - 16 u + 7 = 0

4 u^{2} - 16 u + 7 = 0 : u = \frac{7}{2}, u = \frac{1}{2}

4 u^{2} - 16 u + 7 = 0

Solve with the quadratic formula

4 u^{2} - 16 u + 7 = 0

Quadratic Equation Formula:

For

a = 4, b = - 16, c = 7

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 7 = 12

(- 16)^{2} - 4 \cdot 4 \cdot 7

Apply exponent rule:

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

n

is even

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 4 \cdot 7

Multiply the numbers:

4 \cdot 4 \cdot 7 = 112

= 1 6^{2} - 112

1 6^{2} = 256

= 256 - 112

Subtract the numbers:

256 - 112 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 16 ) \pm 12}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 16 ) + 12}{2 \cdot 4}, u_{2} = \frac{- ( - 16 ) - 12}{2 \cdot 4}

u = \frac{- ( - 16 ) + 12}{2 \cdot 4} : \frac{7}{2}

\frac{- ( - 16 ) + 12}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{16 + 12}{2 \cdot 4}

Add the numbers:

16 + 12 = 28

= \frac{28}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{28}{8}

Cancel the common factor:

4

= \frac{7}{2}

u = \frac{- ( - 16 ) - 12}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 16 ) - 12}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{16 - 12}{2 \cdot 4}

Subtract the numbers:

16 - 12 = 4

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{7}{2}, u = \frac{1}{2}

Substitute back

u = tan (x)

tan (x) = \frac{7}{2}, tan (x) = \frac{1}{2}

tan (x) = \frac{7}{2}, tan (x) = \frac{1}{2}

tan (x) = \frac{7}{2} : x = arctan (\frac{7}{2}) + πn

tan (x) = \frac{7}{2}

Apply trig inverse properties

tan (x) = \frac{7}{2}

General solutions for

tan (x) = \frac{7}{2}

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

x = arctan (\frac{7}{2}) + πn

x = arctan (\frac{7}{2}) + πn

tan (x) = \frac{1}{2} : x = arctan (\frac{1}{2}) + πn

tan (x) = \frac{1}{2}

Apply trig inverse properties

tan (x) = \frac{1}{2}

General solutions for

tan (x) = \frac{1}{2}

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

x = arctan (\frac{1}{2}) + πn

x = arctan (\frac{1}{2}) + πn

Combine all the solutions

x = arctan (\frac{7}{2}) + πn, x = arctan (\frac{1}{2}) + πn

Show solutions in decimal form

x = 1.29249 \dots + πn, x = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(x)-16tan(x)+7=0 ?
The general solution for 4tan^2(x)-16tan(x)+7=0 is x=1.29249…+pin,x=0.46364…+pin