What is the general solution for sin(2x+110)+sin(2x-10)= 1/2 ?

The general solution for sin(2x+110)+sin(2x-10)= 1/2 is x=-10+180n,x=50+180n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2x+110)+sin(2x-10)= 1/2

sin (2 x + 11 0^{\circ}) + sin (2 x - 1 0^{\circ}) = \frac{1}{2}

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

Solution steps

sin (2 x + 11 0^{\circ}) + sin (2 x - 1 0^{\circ}) = \frac{1}{2}

Rewrite using trig identities

sin (\frac{36 x + 90 0 ^{\circ}}{18}) = \frac{1}{2}

General solutions for

sin (\frac{36 x + 90 0 ^{\circ}}{18}) = \frac{1}{2}

\frac{36 x + 90 0 ^{\circ}}{18} = 3 0^{\circ} + 36 0^{\circ} n, \frac{36 x + 90 0 ^{\circ}}{18} = 15 0^{\circ} + 36 0^{\circ} n

Solve

\frac{36 x + 90 0 ^{\circ}}{18} = 3 0^{\circ} + 36 0^{\circ} n : x = - 1 0^{\circ} + 18 0^{\circ} n

Solve

\frac{36 x + 90 0 ^{\circ}}{18} = 15 0^{\circ} + 36 0^{\circ} n : x = 5 0^{\circ} + 18 0^{\circ} n

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

6 cot^{2} (x) - 18 = 0

(2 sin (θ) - 3) (sin (θ) - 1) = 0

20 sin^{2} (x) + cos (x) - 19 = 0

sin^{2} (y) = 1

2 cos (2 θ) + 15 cos (θ) + 5 = 7 cos (θ)

What is the general solution for sin(2x+110)+sin(2x-10)= 1/2 ?
The general solution for sin(2x+110)+sin(2x-10)= 1/2 is x=-10+180n,x=50+180n