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

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

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cos(2x)+3cos(x)+2=0, pi/2 <= x<= pi

cos (2 x) + 3 cos (x) + 2 = 0, \frac{π}{2} \leq x \leq π

x = \frac{2 π}{3}, x = π

Solution steps

cos (2 x) + 3 cos (x) + 2 = 0, \frac{π}{2} \leq x \leq π

Rewrite using trig identities

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

Solve by substitution

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

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

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

Combine all the solutions

x = \frac{2 π}{3}, x = π

2 cos (a) = 3

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

sin^{2} (4 x) = 1

sinh (x) = \frac{2}{3}

cos (α_{2}) = \frac{1}{14}, 0 \leq α_{2} \leq 2 π

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