What is the general solution for sin(3x+10)=cos(x+20) ?

The general solution for sin(3x+10)=cos(x+20) is x=(1080n+180)/(12),x=(900+3240n)/(18)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(3x+10)=cos(x+20)

sin (3 x + 1 0^{\circ}) = cos (x + 2 0^{\circ})

x = \frac{108 0 ^{\circ} n + 18 0 ^{\circ}}{12}, x = \frac{90 0 ^{\circ} + 324 0 ^{\circ} n}{18}

Solution steps

sin (3 x + 1 0^{\circ}) = cos (x + 2 0^{\circ})

Rewrite using trig identities

sin (3 x + 1 0^{\circ}) = sin (9 0^{\circ} - (x + 2 0^{\circ}))

Apply trig inverse properties

3 x + 1 0^{\circ} = 9 0^{\circ} - (x + 2 0^{\circ}) + 36 0^{\circ} n, 3 x + 1 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - (x + 2 0^{\circ})) + 36 0^{\circ} n

3 x + 1 0^{\circ} = 9 0^{\circ} - (x + 2 0^{\circ}) + 36 0^{\circ} n : x = \frac{108 0 ^{\circ} n + 18 0 ^{\circ}}{12}

3 x + 1 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - (x + 2 0^{\circ})) + 36 0^{\circ} n : x = \frac{90 0 ^{\circ} + 324 0 ^{\circ} n}{18}

x = \frac{108 0 ^{\circ} n + 18 0 ^{\circ}}{12}, x = \frac{90 0 ^{\circ} + 324 0 ^{\circ} n}{18}

tan (θ) = \frac{300}{400}

2 cot^{2} (3 x) = 5 csc (3 x) - 4

cos (x) = \frac{- 3}{3}

cos (9 x) = \frac{1}{2}

tan (2 θ) = - 3, 0^{\circ} \leq θ \leq 36 0^{\circ}

What is the general solution for sin(3x+10)=cos(x+20) ?
The general solution for sin(3x+10)=cos(x+20) is x=(1080n+180)/(12),x=(900+3240n)/(18)