What is the general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi ?

The general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi is x=2.15146…,x=-2.15146…+2pi

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3cos^2(x)-2cos(x)-2=0,0<= x<2pi

3 cos^{2} (x) - 2 cos (x) - 2 = 0, 0 \leq x < 2 π

x = 2.15146 \dots, x = - 2.15146 \dots + 2 π

Solution steps

3 cos^{2} (x) - 2 cos (x) - 2 = 0, 0 \leq x < 2 π

Solve by substitution

cos (x) = \frac{1 + 7}{3}, cos (x) = \frac{1 - 7}{3}

cos (x) = \frac{1 + 7}{3}, 0 \leq x < 2 π :

No Solution

cos (x) = \frac{1 - 7}{3}, 0 \leq x < 2 π : x = arccos (\frac{1 - 7}{3}), x = - arccos (\frac{1 - 7}{3}) + 2 π

Combine all the solutions

x = arccos (\frac{1 - 7}{3}), x = - arccos (\frac{1 - 7}{3}) + 2 π

Show solutions in decimal form

x = 2.15146 \dots, x = - 2.15146 \dots + 2 π

sin (x + \frac{π}{4}) + sin (x - \frac{π}{4}) + \frac{6}{2} = 0

tan (θ) = \frac{500}{300}

cos^{2} (x) sin (x) = sin (x), (0, 2 π)

cos (x) = - \frac{3}{7}

cos (x) - sin (7 x) = 0

What is the general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi ?
The general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi is x=2.15146…,x=-2.15146…+2pi