What is the general solution for (1.33)sin(25)=(1.5)sin(θ) ?

The general solution for (1.33)sin(25)=(1.5)sin(θ) is θ=0.38409…+360n,θ=180-0.38409…+360n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1.33)sin(25)=(1.5)sin(θ)

(1.33) sin (2 5^{\circ}) = (1.5) sin (θ)

θ = 0.38409 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.38409 \dots + 36 0^{\circ} n

Solution steps

(1.33) sin (2 5^{\circ}) = (1.5) sin (θ)

Switch sides

1.5 sin (θ) = 1.33 sin (2 5^{\circ})

Divide both sides by

1.5

sin (θ) = \frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}

Apply trig inverse properties

θ = arcsin (\frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 0.38409 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.38409 \dots + 36 0^{\circ} n

cos (y) = - \frac{1}{2}

\frac{2}{1 + sin ( x )} = 3 - 2 sin (x)

x, 4 cos^{2} (x) = 5 + 4 sin (x)

t, y = cos (2 t)

cos (3 x) = \frac{1}{5}

What is the general solution for (1.33)sin(25)=(1.5)sin(θ) ?
The general solution for (1.33)sin(25)=(1.5)sin(θ) is θ=0.38409…+360n,θ=180-0.38409…+360n