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 < 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 < 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 < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 3, 0 \leq x < 2 π :

No Solution

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}

8 - 12 sin^{2} (θ) = 4 cos^{2} (θ)

\frac{sin ( 3 4 ^{\circ} )}{12} = \frac{sin ( x )}{15}

3 - 3 cos (θ) = \frac{3}{2} sec (θ)

18 sec^{2} (x) - 3 tan (x) = 21

sin^{2} (θ) - 2 cos (θ) = 2

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