What is the general solution for cos(2x)+5sin(x)+2=0,0<= x<= 2pi ?

The general solution for cos(2x)+5sin(x)+2=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6

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cos(2x)+5sin(x)+2=0,0<= x<= 2pi

cos (2 x) + 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

cos (2 x) + 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

3 - 2 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

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

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 3, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}

2 cos (4 θ) - 1 = 0

3 cos (x) = 8 tan (x)

1 + 4 sin^{2} (x) - 5 sin (x) + cos (2 x) = 0

cos (x) = \frac{9}{10}

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

What is the general solution for cos(2x)+5sin(x)+2=0,0<= x<= 2pi ?
The general solution for cos(2x)+5sin(x)+2=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6