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

The general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi is x=0.84106…,x=2pi-0.84106…,x=pi

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

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

x = 0.84106 \dots, x = 2 π - 0.84106 \dots, x = π

Solution steps

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

Solve by substitution

cos (x) = \frac{2}{3}, cos (x) = - 1

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

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

Combine all the solutions

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

Show solutions in decimal form

x = 0.84106 \dots, x = 2 π - 0.84106 \dots, x = π

- 1 = tan^{2} (θ)

y, x = - cos (y)

tan (x) = \frac{3}{6}

θ, r = 1 + cos (2 θ)

tan (x) + tan (x + 4 5^{\circ}) = - 2

What is the general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi ?
The general solution for 3cos^2(x)+cos(x)-2=0,0<= x<= 2pi is x=0.84106…,x=2pi-0.84106…,x=pi