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

The general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 is x=-0.46364…+pin,x=0.58800…+pin

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

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

x = - 0.46364 \dots + πn, x = 0.58800 \dots + πn

Solution steps

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

Factor

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

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

Solving each part separately

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

2 sin (x) + cos (x) = 0 : x = arctan (- \frac{1}{2}) + πn

3 sin (x) - 2 cos (x) = 0 : x = arctan (\frac{2}{3}) + πn

Combine all the solutions

x = arctan (- \frac{1}{2}) + πn, x = arctan (\frac{2}{3}) + πn

Show solutions in decimal form

x = - 0.46364 \dots + πn, x = 0.58800 \dots + πn

tan (θ) = \frac{20 ( 9.8 )}{25}

t, 0 = 8 sin (\frac{π t}{12}) + 32

sec^{2} (x) = \frac{1}{sin ^{2} ( x )}

2 sin^{2} (x) + sin (x) - 1 = 0, 0 \leq x < 2 π

arctan (x) = \frac{7.8}{- 4.42}

What is the general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 ?
The general solution for 6sin^2(x)-sin(x)cos(x)-2cos^2(x)=0 is x=-0.46364…+pin,x=0.58800…+pin