What is the general solution for 2cos^2(x)-7cos(x)+3=0,(0,2pi) ?

The general solution for 2cos^2(x)-7cos(x)+3=0,(0,2pi) is x= pi/3 ,x=(5pi)/3

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2cos^2(x)-7cos(x)+3=0,(0,2pi)

2 cos^{2} (x) - 7 cos (x) + 3 = 0, (0, 2 π)

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

2 cos^{2} (x) - 7 cos (x) + 3 = 0, (0, 2 π)

Solve by substitution

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

cos (x) = 3, (0, 2 π) :

No Solution

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

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

cos^{2} (x) - 3 sin^{2} (x) - 1 = 0

tanh (x) + 4 sech (x) = 4

\frac{36}{49} + cos^{2} (θ) = 1

sin (x) + \frac{3}{2} = 0

3 cos^{2} (x) + 10 cos (x) + 3 = 0

What is the general solution for 2cos^2(x)-7cos(x)+3=0,(0,2pi) ?
The general solution for 2cos^2(x)-7cos(x)+3=0,(0,2pi) is x= pi/3 ,x=(5pi)/3