What is the general solution for 2csc(θ)=4cos(θ) ?

The general solution for 2csc(θ)=4cos(θ) is θ= pi/4+pin

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Popular >

2csc(θ)=4cos(θ)

Solution

2 csc (θ) = 4 cos (θ)

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

2 csc (θ) = 4 cos (θ)

Subtract

4 cos (θ)

from both sides

2 csc (θ) - 4 cos (θ) = 0

Express with sin, cos

2 csc (θ) - 4 cos (θ)

Use the basic trigonometric identity:

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

= 2 \cdot \frac{1}{sin ( θ )} - 4 cos (θ)

Simplify

2 \cdot \frac{1}{sin ( θ )} - 4 cos (θ) : \frac{2 - 4 cos ( θ ) sin ( θ )}{sin ( θ )}

2 \cdot \frac{1}{sin ( θ )} - 4 cos (θ)

2 \cdot \frac{1}{sin ( θ )} = \frac{2}{sin ( θ )}

2 \cdot \frac{1}{sin ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 2}{sin ( θ )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sin ( θ )}

= \frac{2}{sin ( θ )} - 4 cos (θ)

Convert element to fraction:

4 cos (θ) = \frac{4 cos ( θ ) sin ( θ )}{sin ( θ )}

= \frac{2}{sin ( θ )} - \frac{4 cos ( θ ) sin ( θ )}{sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 - 4 cos ( θ ) sin ( θ )}{sin ( θ )}

= \frac{2 - 4 cos ( θ ) sin ( θ )}{sin ( θ )}

\frac{2 - 4 cos ( θ ) sin ( θ )}{sin ( θ )} = 0

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

2 - 4 cos (θ) sin (θ) = 0

Rewrite using trig identities

2 - 4 cos (θ) sin (θ)

Use the Double Angle identity:

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

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

= 2 - 4 \cdot \frac{sin ( 2 θ )}{2}

2 - 4 \cdot \frac{sin ( 2 θ )}{2} = 0

4 \cdot \frac{sin ( 2 θ )}{2} = 2 sin (2 θ)

4 \cdot \frac{sin ( 2 θ )}{2}

Multiply fractions:

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

= \frac{sin ( 2 θ ) \cdot 4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 sin (2 θ)

2 - 2 sin (2 θ) = 0

Move

2

to the right side

2 - 2 sin (2 θ) = 0

Subtract

2

from both sides

2 - 2 sin (2 θ) - 2 = 0 - 2

Simplify

- 2 sin (2 θ) = - 2

- 2 sin (2 θ) = - 2

Divide both sides by

- 2

- 2 sin (2 θ) = - 2

Divide both sides by

- 2

\frac{- 2 sin ( 2 θ )}{- 2} = \frac{- 2}{- 2}

Simplify

sin (2 θ) = 1

sin (2 θ) = 1

General solutions for

sin (2 θ) = 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}

2 θ = \frac{π}{2} + 2 πn

2 θ = \frac{π}{2} + 2 πn

Solve

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Popular Examples

6/5 arccos(y/6)=pi

\frac{6}{5} arccos (\frac{y}{6}) = π

cos(x)=(154)/(212.85)

cos (x) = \frac{154}{212.85}

4sin^2(x)-2=0,0<= x<= 2pi

4 sin^{2} (x) - 2 = 0, 0 \leq x \leq 2 π

csc^2(θ)=7cot(θ)+9

csc^{2} (θ) = 7 cot (θ) + 9

cos^2(x)+6cos(x)=-3,y=3

cos^{2} (x) + 6 cos (x) = - 3, y = 3

Frequently Asked Questions (FAQ)

What is the general solution for 2csc(θ)=4cos(θ) ?
The general solution for 2csc(θ)=4cos(θ) is θ= pi/4+pin