Solving for trigonometric equation.

[Chadochocinco]

3csc^25x=-4 or 3 x csc^2 x 5x= -4 (3csc squared times 5x equals negative 4.)

Note: the 5x is being multiplied to the 3csc. The trouble I am having is that i cannot square the other side of the equation to simplify or solve using quadratic formula because there is a negative four, and I am unable to factor it out because it isn't possible, or at least i don't know how for this type of problem.

 

[galactus]

Are you sure it isn't $\displaystyle 3csc^{2}(5x)=-4$?.

That would seem more likely than $\displaystyle 3\cdot 5x\cdot csc^{2}(x)=-4\rightarrow 15x\cdot csc^{2}(x)=-4$

 

[HallsofIvy]

Be careful with terminology. In "$\displaystyle csc^2(5x)$" you are NOT "multiplying" 5x by csc. 5x is the argument in the function $\displaystyle csc^2( )$.

Of course, if really is $\displaystyle 3csc^2(5x)= -4$ there is no real number solution because a square cannot be negative.

 

[Chadochocinco]

Be careful with terminology. In "$\displaystyle csc^2(5x)$" you are NOT "multiplying" 5x by csc. 5x is the argument in the function $\displaystyle csc^2( )$.

Of course, if really is $\displaystyle 3csc^2(5x)= -4$ there is no real number solution because a square cannot be negative.

You're right sorry. So in that case would it still be no solution because there is a negative number, therefore it's undefined?