What is the general solution for sin(2θ)= 1/3 ?

The general solution for sin(2θ)= 1/3 is θ=(0.33983…)/2+pin,θ= pi/2-(0.33983…)/2+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2θ)= 1/3

sin (2 θ) = \frac{1}{3}

θ = \frac{0.33983 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

Degrees

θ = 9.73561 \dots^{\circ} + 18 0^{\circ} n, θ = 80.26438 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 θ) = \frac{1}{3}

Apply trig inverse properties

sin (2 θ) = \frac{1}{3}

General solutions for

sin (2 θ) = \frac{1}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 θ = arcsin (\frac{1}{3}) + 2 πn, 2 θ = π - arcsin (\frac{1}{3}) + 2 πn

2 θ = arcsin (\frac{1}{3}) + 2 πn, 2 θ = π - arcsin (\frac{1}{3}) + 2 πn

Solve

2 θ = arcsin (\frac{1}{3}) + 2 πn : θ = \frac{arcsin ( \frac{1}{3} )}{2} + πn

2 θ = arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 θ = arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arcsin ( \frac{1}{3} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{3} )}{2} + πn

Solve

2 θ = π - arcsin (\frac{1}{3}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

2 θ = π - arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 θ = π - arcsin (\frac{1}{3}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{3} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.33983 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.33983 \dots}{2} + πn

cos^{2} (θ) = 1 - sin (θ)

1 sin (3 0^{\circ}) = 1.495 sin (x)

e^{2 c o s (x) - 1} = 1

csc (θ) = - \frac{37}{12}, \frac{3 π}{2} \leq θ \leq 2 π

\frac{3}{2} = 2 \cdot 25 sin (\frac{θ}{2})

What is the general solution for sin(2θ)= 1/3 ?
The general solution for sin(2θ)= 1/3 is θ=(0.33983…)/2+pin,θ= pi/2-(0.33983…)/2+pin