What is the general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 ?

The general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 is x=0.32175…+pin,x=-1.24904…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x) = 0

x = 0.32175 \dots + πn, x = - 1.24904 \dots + πn

Solution steps

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x) = 0

Factor

6 sin^{2} (x) + 16 sin (x) cos (x) - 6 cos^{2} (x) : 2 (3 sin (x) - cos (x)) (sin (x) + 3 cos (x))

2 (3 sin (x) - cos (x)) (sin (x) + 3 cos (x)) = 0

Solving each part separately

3 sin (x) - cos (x) = 0 or sin (x) + 3 cos (x) = 0

3 sin (x) - cos (x) = 0 : x = arctan (\frac{1}{3}) + πn

sin (x) + 3 cos (x) = 0 : x = arctan (- 3) + πn

Combine all the solutions

x = arctan (\frac{1}{3}) + πn, x = arctan (- 3) + πn

Show solutions in decimal form

x = 0.32175 \dots + πn, x = - 1.24904 \dots + πn

sec^{2} (t) = 0

x, t = tan (\frac{x}{2})

sin^{2} (x) = 8 cos^{2} (x)

sin (x) + cos (x) = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

sin (θ) = - 0.342

What is the general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 ?
The general solution for 6sin^2(x)+16sin(x)cos(x)-6cos^2(x)=0 is x=0.32175…+pin,x=-1.24904…+pin