What is the general solution for sin(x)+4csc(x)+5=0,0<= x<= 2pi ?

The general solution for sin(x)+4csc(x)+5=0,0<= x<= 2pi is x=(3pi)/2

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sin(x)+4csc(x)+5=0,0<= x<= 2pi

sin (x) + 4 csc (x) + 5 = 0, 0 \leq x \leq 2 π

x = \frac{3 π}{2}

Solution steps

sin (x) + 4 csc (x) + 5 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

5 + \frac{1}{csc ( x )} + 4 csc (x) = 0

Solve by substitution

csc (x) = - \frac{1}{4}, csc (x) = - 1

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

No Solution

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

Combine all the solutions

x = \frac{3 π}{2}

r = a sin (3 x)

sin (x) = \frac{18}{25}

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

x, cos (x) = \frac{1}{4}

cos^{2} (t) = 0

What is the general solution for sin(x)+4csc(x)+5=0,0<= x<= 2pi ?
The general solution for sin(x)+4csc(x)+5=0,0<= x<= 2pi is x=(3pi)/2