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

Solution

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

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π, x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

Solution steps

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

Solve by substitution

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

cos (3 x) = 1, 0 \leq x \leq 2 π : x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

cos (3 x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

Combine all the solutions

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π, x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

x, z = y^{s i n (x)}

t, 10 = 14 + 8 sin (\frac{π t}{12})

cos (4 x) cos (x - 1) = 0

cos (0.5 t + 0.464) = 1

2 - 2 sin (θ) = 3