What is the general solution for 3sec^2(u)+7tan(u)=3 ?

The general solution for 3sec^2(u)+7tan(u)=3 is u=pin,u=-1.16590…+pin

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Popular >

3sec^2(u)+7tan(u)=3

Solution

3 sec^{2} (u) + 7 tan (u) = 3

Solution

u = πn, u = - 1.16590 \dots + πn

Degrees

u = 0^{\circ} + 18 0^{\circ} n, u = - 66.80140 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 sec^{2} (u) + 7 tan (u) = 3

Subtract

3

from both sides

3 sec^{2} (u) + 7 tan (u) - 3 = 0

Rewrite using trig identities

- 3 + 3 sec^{2} (u) + 7 tan (u)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= - 3 + 3 (tan^{2} (u) + 1) + 7 tan (u)

Simplify

- 3 + 3 (tan^{2} (u) + 1) + 7 tan (u) : 3 tan^{2} (u) + 7 tan (u)

- 3 + 3 (tan^{2} (u) + 1) + 7 tan (u)

Expand

3 (tan^{2} (u) + 1) : 3 tan^{2} (u) + 3

3 (tan^{2} (u) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = tan^{2} (u), c = 1

= 3 tan^{2} (u) + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3 tan^{2} (u) + 3

= - 3 + 3 tan^{2} (u) + 3 + 7 tan (u)

Simplify

- 3 + 3 tan^{2} (u) + 3 + 7 tan (u) : 3 tan^{2} (u) + 7 tan (u)

- 3 + 3 tan^{2} (u) + 3 + 7 tan (u)

Group like terms

= 3 tan^{2} (u) + 7 tan (u) - 3 + 3

- 3 + 3 = 0

= 3 tan^{2} (u) + 7 tan (u)

= 3 tan^{2} (u) + 7 tan (u)

= 3 tan^{2} (u) + 7 tan (u)

3 tan^{2} (u) + 7 tan (u) = 0

Solve by substitution

3 tan^{2} (u) + 7 tan (u) = 0

Let:

tan (u) = v

3 v^{2} + 7 v = 0

3 v^{2} + 7 v = 0 : v = 0, v = - \frac{7}{3}

3 v^{2} + 7 v = 0

Solve with the quadratic formula

3 v^{2} + 7 v = 0

Quadratic Equation Formula:

For

a = 3, b = 7, c = 0

v_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

v_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

7^{2} - 4 \cdot 3 \cdot 0 = 7

7^{2} - 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

Apply radical rule:

assuming

a \geq 0

= 7

v_{1, 2} = \frac{- 7 \pm 7}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- 7 + 7}{2 \cdot 3}, v_{2} = \frac{- 7 - 7}{2 \cdot 3}

v = \frac{- 7 + 7}{2 \cdot 3} : 0

\frac{- 7 + 7}{2 \cdot 3}

Add/Subtract the numbers:

- 7 + 7 = 0

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{6}

Apply rule

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

= 0

v = \frac{- 7 - 7}{2 \cdot 3} : - \frac{7}{3}

\frac{- 7 - 7}{2 \cdot 3}

Subtract the numbers:

- 7 - 7 = - 14

= \frac{- 14}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 14}{6}

Apply the fraction rule:

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

= - \frac{14}{6}

Cancel the common factor:

2

= - \frac{7}{3}

The solutions to the quadratic equation are:

v = 0, v = - \frac{7}{3}

Substitute back

v = tan (u)

tan (u) = 0, tan (u) = - \frac{7}{3}

tan (u) = 0, tan (u) = - \frac{7}{3}

tan (u) = 0 : u = πn

tan (u) = 0

General solutions for

tan (u) = 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}

u = 0 + πn

u = 0 + πn

Solve

u = 0 + πn : u = πn

u = 0 + πn

0 + πn = πn

u = πn

u = πn

tan (u) = - \frac{7}{3} : u = arctan (- \frac{7}{3}) + πn

tan (u) = - \frac{7}{3}

Apply trig inverse properties

tan (u) = - \frac{7}{3}

General solutions for

tan (u) = - \frac{7}{3}

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

u = arctan (- \frac{7}{3}) + πn

u = arctan (- \frac{7}{3}) + πn

Combine all the solutions

u = πn, u = arctan (- \frac{7}{3}) + πn

Show solutions in decimal form

u = πn, u = - 1.16590 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sec^2(u)+7tan(u)=3 ?
The general solution for 3sec^2(u)+7tan(u)=3 is u=pin,u=-1.16590…+pin