What is the general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 ?

Q: What is the general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 ?

The general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 is θ=30,θ=60

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Popular Trigonometry >

4sin(θ)=sqrt(3)sec(θ),0<= θ<180

Solution

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

Solution

θ = 3 0^{\circ}, θ = 6 0^{\circ}

Radians

θ = \frac{π}{6}, θ = \frac{π}{3}

Solution steps

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

Subtract

3 sec (θ)

from both sides

4 sin (θ) - 3 sec (θ) = 0

Express with sin, cos

4 sin (θ) - sec (θ) 3

Use the basic trigonometric identity:

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

= 4 sin (θ) - \frac{1}{cos ( θ )} 3

Simplify

4 sin (θ) - \frac{1}{cos ( θ )} 3 : \frac{4 sin ( θ ) cos ( θ ) - 3}{cos ( θ )}

4 sin (θ) - \frac{1}{cos ( θ )} 3

\frac{1}{cos ( θ )} 3 = \frac{3}{cos ( θ )}

\frac{1}{cos ( θ )} 3

Multiply fractions:

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

= \frac{1 \cdot 3}{cos ( θ )}

Multiply:

1 \cdot 3 = 3

= \frac{3}{cos ( θ )}

= 4 sin (θ) - \frac{3}{cos ( θ )}

Convert element to fraction:

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

= \frac{4 sin ( θ ) cos ( θ )}{cos ( θ )} - \frac{3}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{4 sin ( θ ) cos ( θ ) - 3}{cos ( θ )}

= \frac{4 sin ( θ ) cos ( θ ) - 3}{cos ( θ )}

\frac{- 3 + 4 cos ( θ ) sin ( θ )}{cos ( θ )} = 0

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

- 3 + 4 cos (θ) sin (θ) = 0

Rewrite using trig identities

- 3 + 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}

= - 3 + 4 \cdot \frac{sin ( 2 θ )}{2}

- 3 + 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)

- 3 + 2 sin (2 θ) = 0

Move

3

to the right side

- 3 + 2 sin (2 θ) = 0

Add

3

to both sides

- 3 + 2 sin (2 θ) + 3 = 0 + 3

Simplify

2 sin (2 θ) = 3

2 sin (2 θ) = 3

Divide both sides by

2

2 sin (2 θ) = 3

Divide both sides by

2

\frac{2 sin ( 2 θ )}{2} = \frac{3}{2}

Simplify

sin (2 θ) = \frac{3}{2}

sin (2 θ) = \frac{3}{2}

General solutions for

sin (2 θ) = \frac{3}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 6 0^{\circ} + 36 0^{\circ} n, 2 θ = 12 0^{\circ} + 36 0^{\circ} n

2 θ = 6 0^{\circ} + 36 0^{\circ} n, 2 θ = 12 0^{\circ} + 36 0^{\circ} n

Solve

2 θ = 6 0^{\circ} + 36 0^{\circ} n : θ = 3 0^{\circ} + 18 0^{\circ} n

2 θ = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 3 0^{\circ} + 18 0^{\circ} n

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

\frac{6 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 3 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

Solve

2 θ = 12 0^{\circ} + 36 0^{\circ} n : θ = 6 0^{\circ} + 18 0^{\circ} n

2 θ = 12 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 θ = 12 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 6 0^{\circ} + 18 0^{\circ} n

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{12 0 ^{\circ}}{2} = 6 0^{\circ}

\frac{12 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{36 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 6 0^{\circ}

Cancel the common factor:

2

= 6 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 6 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq θ < 18 0^{\circ}

θ = 3 0^{\circ}, θ = 6 0^{\circ}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 ?
The general solution for 4sin(θ)=sqrt(3)sec(θ),0<= θ<180 is θ=30,θ=60