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

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

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

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

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

Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

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

cos^{2} (a) = \frac{1}{2}

2 sin^{2} (x) + 3 sin (x) + 1 = 0, 0 \leq x < 2 π

cos (15 θ) = 0

sin (t) cos (t) = 0

6 sin (θ) = 0

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