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

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

The general solution for (sin(2x))/(cos(x))=2,0<= x<= pi is No Solution for x\in\mathbb{R}

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(sin(2x))/(cos(x))=2,0<= x<= pi

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

2

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

sin (2 x) - 2 cos (x) = 0

Rewrite using trig identities

sin (2 x) - 2 cos (x)

Use the Double Angle identity:

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

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

- 2 cos (x) + 2 cos (x) sin (x) = 0

Factor

- 2 cos (x) + 2 cos (x) sin (x) : 2 cos (x) (sin (x) - 1)

- 2 cos (x) + 2 cos (x) sin (x)

Factor out common term

2 cos (x)

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

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

Solving each part separately

cos (x) = 0 or sin (x) - 1 = 0

cos (x) = 0, 0 \leq x \leq π : x = \frac{π}{2}

cos (x) = 0, 0 \leq x \leq π

General solutions for

cos (x) = 0

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

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

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

Solutions for the range

0 \leq x \leq π

x = \frac{π}{2}

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

sin (x) - 1 = 0, 0 \leq x \leq π

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

sin (x) - 1 + 1 = 0 + 1

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x \leq π

x = \frac{π}{2}

Combine all the solutions

x = \frac{π}{2}

Since the equation is undefined for:

\frac{π}{2}

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(2x))/(cos(x))=2,0<= x<= pi ?
The general solution for (sin(2x))/(cos(x))=2,0<= x<= pi is No Solution for x\in\mathbb{R}