What is the general solution for 2cos(x)+cos(2x)*2=0 ?

The general solution for 2cos(x)+cos(2x)*2=0 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin

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2cos(x)+cos(2x)*2=0

2 cos (x) + cos (2 x) \cdot 2 = 0

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

Solution steps

2 cos (x) + cos (2 x) \cdot 2 = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (x)) \cdot 2 + 2 cos (x) = 0

Solve by substitution

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

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

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

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

3 sin^{2} (t) - 7 sin (t) - 6 = 0

sin (s) = - \frac{2}{2}

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ < 2 π

950 sin (\frac{π}{6} (7 - x)) + 1650 = 2500

sin (θ) = 12

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