What is the general solution for tan(x)= 1/(sin(2x)),0<= x<= pi ?

Q: What is the general solution for tan(x)= 1/(sin(2x)),0<= x<= pi ?

The general solution for tan(x)= 1/(sin(2x)),0<= x<= pi is x= pi/4 ,x=(3pi)/4

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tan(x)= 1/(sin(2x)),0<= x<= pi

Solution

tan (x) = \frac{1}{sin ( 2 x )}, 0 \leq x \leq π

Solution

x = \frac{π}{4}, x = \frac{3 π}{4}

Degrees

x = 4 5^{\circ}, x = 13 5^{\circ}

Solution steps

tan (x) = \frac{1}{sin ( 2 x )}, 0 \leq x \leq π

Subtract

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

from both sides

tan (x) - \frac{1}{sin ( 2 x )} = 0

Simplify

tan (x) - \frac{1}{sin ( 2 x )} : \frac{tan ( x ) sin ( 2 x ) - 1}{sin ( 2 x )}

tan (x) - \frac{1}{sin ( 2 x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

\frac{tan ( x ) sin ( 2 x ) - 1}{sin ( 2 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

- 1 + sin (2 x) tan (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= - 1 + 2 sin (x) cos (x) tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + 2 cos (x) sin (x) \frac{sin ( x )}{cos ( x )}

2 cos (x) sin (x) \frac{sin ( x )}{cos ( x )} = 2 sin^{2} (x)

2 cos (x) sin (x) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) \cdot 2 cos ( x ) sin ( x )}{cos ( x )}

Cancel the common factor:

cos (x)

= sin (x) \cdot 2 sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= 2 sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 2 sin^{2} (x)

= - 1 + 2 sin^{2} (x)

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

Solve by substitution

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

Let:

sin (x) = u

- 1 + 2 u^{2} = 0

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

- 1 + 2 u^{2} = 0

Move

1

to the right side

- 1 + 2 u^{2} = 0

Add

1

to both sides

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

Simplify

2 u^{2} = 1

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, 0 \leq x \leq π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, 0 \leq x \leq π

General solutions for

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solutions for the range

0 \leq x \leq π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq π :

No Solution

sin (x) = - \frac{1}{2}, 0 \leq x \leq π

General solutions for

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x \leq π

No Solution

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)= 1/(sin(2x)),0<= x<= pi ?
The general solution for tan(x)= 1/(sin(2x)),0<= x<= pi is x= pi/4 ,x=(3pi)/4