What is the general solution for (sec(3x)-sqrt(2))(tan(2x)-1)=0 ?

The general solution for (sec(3x)-sqrt(2))(tan(2x)-1)=0 is x= pi/8+(pin)/2

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(sec(3x)-sqrt(2))(tan(2x)-1)=0

Solution

(sec (3 x) - 2) (tan (2 x) - 1) = 0

Solution

x = \frac{π}{8} + \frac{πn}{2}

Degrees

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

Solution steps

(sec (3 x) - 2) (tan (2 x) - 1) = 0

Solving each part separately

sec (3 x) - 2 = 0 or tan (2 x) - 1 = 0

sec (3 x) - 2 = 0 : x = \frac{π}{12} + \frac{2 πn}{3}, x = \frac{7 π}{12} + \frac{2 πn}{3}

sec (3 x) - 2 = 0

Move

2

to the right side

sec (3 x) - 2 = 0

Add

2

to both sides

sec (3 x) - 2 + 2 = 0 + 2

Simplify

sec (3 x) = 2

sec (3 x) = 2

General solutions for

sec (3 x) = 2

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

3 x = \frac{π}{4} + 2 πn, 3 x = \frac{7 π}{4} + 2 πn

3 x = \frac{π}{4} + 2 πn, 3 x = \frac{7 π}{4} + 2 πn

Solve

3 x = \frac{π}{4} + 2 πn : x = \frac{π}{12} + \frac{2 πn}{3}

3 x = \frac{π}{4} + 2 πn

Divide both sides by

3

3 x = \frac{π}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{4}}{3} + \frac{2 πn}{3} : \frac{π}{12} + \frac{2 πn}{3}

\frac{\frac{π}{4}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{2 πn}{3}

x = \frac{π}{12} + \frac{2 πn}{3}

x = \frac{π}{12} + \frac{2 πn}{3}

x = \frac{π}{12} + \frac{2 πn}{3}

Solve

3 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{12} + \frac{2 πn}{3}

3 x = \frac{7 π}{4} + 2 πn

Divide both sides by

3

3 x = \frac{7 π}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3} : \frac{7 π}{12} + \frac{2 πn}{3}

\frac{\frac{7 π}{4}}{3} + \frac{2 πn}{3}

\frac{\frac{7 π}{4}}{3} = \frac{7 π}{12}

\frac{\frac{7 π}{4}}{3}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{7 π}{12}

= \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{π}{12} + \frac{2 πn}{3}, x = \frac{7 π}{12} + \frac{2 πn}{3}

tan (2 x) - 1 = 0 : x = \frac{π}{8} + \frac{πn}{2}

tan (2 x) - 1 = 0

Move

1

to the right side

tan (2 x) - 1 = 0

Add

1

to both sides

tan (2 x) - 1 + 1 = 0 + 1

Simplify

tan (2 x) = 1

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}

Combine all the solutions

x = \frac{π}{12} + \frac{2 πn}{3}, x = \frac{7 π}{12} + \frac{2 πn}{3}, x = \frac{π}{8} + \frac{πn}{2}

Since the equation is undefined for:

\frac{π}{12} + \frac{2 πn}{3}, \frac{7 π}{12} + \frac{2 πn}{3}

x = \frac{π}{8} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for (sec(3x)-sqrt(2))(tan(2x)-1)=0 ?
The general solution for (sec(3x)-sqrt(2))(tan(2x)-1)=0 is x= pi/8+(pin)/2