What is the general solution for 2sec^2(2x)=3tan(2x)+1 ?

The general solution for 2sec^2(2x)=3tan(2x)+1 is x= pi/8+(pin)/2 ,x=(0.46364…)/2+(pin)/2

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2sec^2(2x)=3tan(2x)+1

Solution

2 sec^{2} (2 x) = 3 tan (2 x) + 1

Solution

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{0.46364 \dots}{2} + \frac{πn}{2}

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n, x = 13.28252 \dots^{\circ} + 9 0^{\circ} n

Solution steps

2 sec^{2} (2 x) = 3 tan (2 x) + 1

Subtract

3 tan (2 x) + 1

from both sides

2 sec^{2} (2 x) - 3 tan (2 x) - 1 = 0

Rewrite using trig identities

- 1 + 2 sec^{2} (2 x) - 3 tan (2 x)

Use the Pythagorean identity:

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

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

Simplify

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

- 1 + 2 (tan^{2} (2 x) + 1) - 3 tan (2 x)

Expand

2 (tan^{2} (2 x) + 1) : 2 tan^{2} (2 x) + 2

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

- 1 + 2 tan^{2} (2 x) + 2 - 3 tan (2 x)

Group like terms

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

Add/Subtract the numbers:

- 1 + 2 = 1

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

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

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

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

Solve by substitution

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

Let:

tan (2 x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

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

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

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

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

Apply exponent rule:

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

n

is even

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

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 3 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- ( - 3 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = tan (2 x)

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

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

tan (2 x) = 1 : x = \frac{π}{8} + \frac{πn}{2}

tan (2 x) = 1

General solutions for

tan (2 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}

2 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

Solve

2 x = \frac{π}{4} + πn : x = \frac{π}{8} + \frac{πn}{2}

2 x = \frac{π}{4} + πn

Divide both sides by

2

2 x = \frac{π}{4} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

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

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

Apply trig inverse properties

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

General solutions for

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

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

2 x = arctan (\frac{1}{2}) + πn

2 x = arctan (\frac{1}{2}) + πn

Solve

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

2 x = arctan (\frac{1}{2}) + πn

Divide both sides by

2

2 x = arctan (\frac{1}{2}) + πn

Divide both sides by

2

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

Simplify

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

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

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

Combine all the solutions

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

Show solutions in decimal form

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{0.46364 \dots}{2} + \frac{πn}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(2x)=3tan(2x)+1 ?
The general solution for 2sec^2(2x)=3tan(2x)+1 is x= pi/8+(pin)/2 ,x=(0.46364…)/2+(pin)/2