What is the general solution for sin(θ)=cos(130) ?

The general solution for sin(θ)=cos(130) is θ=360n-40,θ=180+40+360n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)=cos(130)

sin (θ) = cos (13 0^{\circ})

θ = 36 0^{\circ} n - 4 0^{\circ}, θ = 18 0^{\circ} + 4 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = cos (13 0^{\circ})

Rewrite using trig identities

sin (θ) = sin (9 0^{\circ} - 13 0^{\circ})

Apply trig inverse properties

θ = 9 0^{\circ} - 13 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} - (9 0^{\circ} - 13 0^{\circ}) + 36 0^{\circ} n

θ = 9 0^{\circ} - 13 0^{\circ} + 36 0^{\circ} n : θ = 36 0^{\circ} n - 4 0^{\circ}

θ = 18 0^{\circ} - (9 0^{\circ} - 13 0^{\circ}) + 36 0^{\circ} n : θ = 18 0^{\circ} + 4 0^{\circ} + 36 0^{\circ} n

θ = 36 0^{\circ} n - 4 0^{\circ}, θ = 18 0^{\circ} + 4 0^{\circ} + 36 0^{\circ} n

2 sin^{2} (θ) - sin (θ) - 2 = 0

3 cot (x) + 2 = 5

sec^{2} (x) - tan (x) - 1 = 0

2 cos (2 x + \frac{π}{6}) + 3 = 2, 0 < x < 2 π

cos (x) - 3 sin (x) = 2

What is the general solution for sin(θ)=cos(130) ?
The general solution for sin(θ)=cos(130) is θ=360n-40,θ=180+40+360n