What is the general solution for 4sin(x)-sec(x)=0 ?

The general solution for 4sin(x)-sec(x)=0 is x= pi/(12)+pin,x=(5pi)/(12)+pin

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4sin(x)-sec(x)=0

Solution

4 sin (x) - sec (x) = 0

Solution

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Degrees

x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n

Solution steps

4 sin (x) - sec (x) = 0

Express with sin, cos

- sec (x) + 4 sin (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= - \frac{1}{cos ( x )} + 4 sin (x)

Simplify

- \frac{1}{cos ( x )} + 4 sin (x) : \frac{- 1 + 4 sin ( x ) cos ( x )}{cos ( x )}

- \frac{1}{cos ( x )} + 4 sin (x)

Convert element to fraction:

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

= - \frac{1}{cos ( x )} + \frac{4 sin ( x ) 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{- 1 + 4 sin ( x ) cos ( x )}{cos ( x )}

= \frac{- 1 + 4 sin ( x ) cos ( x )}{cos ( x )}

\frac{- 1 + 4 cos ( x ) sin ( x )}{cos ( x )} = 0

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

- 1 + 4 cos (x) sin (x) = 0

Rewrite using trig identities

- 1 + 4 cos (x) sin (x)

Use the Double Angle identity:

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

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

= - 1 + 4 \cdot \frac{sin ( 2 x )}{2}

- 1 + 4 \cdot \frac{sin ( 2 x )}{2} = 0

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

4 \cdot \frac{sin ( 2 x )}{2}

Multiply fractions:

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

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

Divide the numbers:

\frac{4}{2} = 2

= 2 sin (2 x)

- 1 + 2 sin (2 x) = 0

Move

1

to the right side

- 1 + 2 sin (2 x) = 0

Add

1

to both sides

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

Simplify

2 sin (2 x) = 1

2 sin (2 x) = 1

Divide both sides by

2

2 sin (2 x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (2 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}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Popular Examples

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

cos(x)=-1/2 ,0<x<2pi

cos (x) = - \frac{1}{2}, 0 < x < 2 π

cot(θ)=cos(θ)

cot (θ) = cos (θ)

3sin^2(θ)-cos(θ)+1=0

3 sin^{2} (θ) - cos (θ) + 1 = 0

2-2cos^2(x)=0

2 - 2 cos^{2} (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 4sin(x)-sec(x)=0 ?
The general solution for 4sin(x)-sec(x)=0 is x= pi/(12)+pin,x=(5pi)/(12)+pin