What is the general solution for 2cos^2(t)-3cos(t)=-1,0<= t<= 2pi ?

The general solution for 2cos^2(t)-3cos(t)=-1,0<= t<= 2pi is t=0,t=2pi,t= pi/3 ,t=(5pi)/3

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2cos^2(t)-3cos(t)=-1,0<= t<= 2pi

2 cos^{2} (t) - 3 cos (t) = - 1, 0 \leq t \leq 2 π

t = 0, t = 2 π, t = \frac{π}{3}, t = \frac{5 π}{3}

Solution steps

2 cos^{2} (t) - 3 cos (t) = - 1, 0 \leq t \leq 2 π

Solve by substitution

cos (t) = 1, cos (t) = \frac{1}{2}

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

cos (t) = \frac{1}{2}, 0 \leq t \leq 2 π : t = \frac{π}{3}, t = \frac{5 π}{3}

Combine all the solutions

t = 0, t = 2 π, t = \frac{π}{3}, t = \frac{5 π}{3}

4 cos (4 x) = 3

tan (y) = - 2

sin (θ) = - \frac{2}{2}

sin^{2} (x) + sin (x) \cdot cos (x) + cos^{2} (x) = 10

x, sin (x) = \frac{2}{3}

What is the general solution for 2cos^2(t)-3cos(t)=-1,0<= t<= 2pi ?
The general solution for 2cos^2(t)-3cos(t)=-1,0<= t<= 2pi is t=0,t=2pi,t= pi/3 ,t=(5pi)/3