What is the general solution for 2cos^2(x)+7cos(x)-4=0 ?

The general solution for 2cos^2(x)+7cos(x)-4=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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2cos^2(x)+7cos(x)-4=0

2 cos^{2} (x) + 7 cos (x) - 4 = 0

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

2 cos^{2} (x) + 7 cos (x) - 4 = 0

Solve by substitution

cos (x) = \frac{1}{2}, cos (x) = - 4

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

cos (x) = - 4 :

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cot (2 θ) = \frac{3}{3}

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

- cos (θ) + 1 = 2 cos (θ) + 3

cot^{2} (x) + cot (x) = 0

3 cos (2 θ) + 9 cos (θ) - 4 = - 2 cos (θ)

What is the general solution for 2cos^2(x)+7cos(x)-4=0 ?
The general solution for 2cos^2(x)+7cos(x)-4=0 is x= pi/3+2pin,x=(5pi)/3+2pin