What is the general solution for 4sin(2x)+csc(2x)=4 ?

The general solution for 4sin(2x)+csc(2x)=4 is x= pi/(12)+pin,x=(5pi)/(12)+pin

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4sin(2x)+csc(2x)=4

4 sin (2 x) + csc (2 x) = 4

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solution steps

4 sin (2 x) + csc (2 x) = 4

Subtract

4

from both sides

4 sin (2 x) + csc (2 x) - 4 = 0

Rewrite using trig identities

- 4 + csc (2 x) + \frac{4}{csc ( 2 x )} = 0

Solve by substitution

csc (2 x) = 2

csc (2 x) = 2 : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Combine all the solutions

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

cot (θ) = \frac{9}{40}

sin (a) = - \frac{1}{2}

5 cos (x) - 2 sin (x) = 0.5

sin (x) = \frac{6}{15}

cot (θ) = 1.4

What is the general solution for 4sin(2x)+csc(2x)=4 ?
The general solution for 4sin(2x)+csc(2x)=4 is x= pi/(12)+pin,x=(5pi)/(12)+pin