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How do you find the general solutions for #cos2xcscx-2cos2x=0#?

Answer 1

Use algebra

#cos 2x csc x - 2 cos 2x = 0# #cos 2x (csc x - 2) = 0# Solution A: #cos 2x = 0 \Rightarrow 2x = pmpi/2 + kpi \Rightarrow x = pi/4 (2kpm1), k in ZZ# Solution B: #1/cos x = 2 \Rightarrow cos x = 1/2 \Rightarrow x = pmpi/3 + 2kpi = pi/3 (6k+-1), k in ZZ# #therefore x = pi/4 (2kpm1)# or #x = pi/3 (6k+-1), k in ZZ#

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Answer 2

Kai Anderson

To find the general solutions for ( \cos(2x)\csc(x) - 2\cos(2x) = 0 ):

Rewrite ( \csc(x) ) as ( \frac{1}{\sin(x)} ).
Factor out ( \cos(2x) ) from the equation.
Set each factor equal to zero and solve for ( x ).
Solve the resulting equations for ( x ).

The steps are as follows:

Rewrite the equation as ( \cos(2x)\left(\frac{1}{\sin(x)} - 2\right) = 0 ).
Set each factor equal to zero: ( \cos(2x) = 0 ) and ( \frac{1}{\sin(x)} - 2 = 0 ).
Solve ( \cos(2x) = 0 ) for ( x ): ( 2x = \frac{\pi}{2} + n\pi ) and ( 2x = \frac{3\pi}{2} + n\pi ), where ( n ) is an integer.
Solve ( \frac{1}{\sin(x)} - 2 = 0 ) for ( x ): ( \sin(x) = \frac{1}{2} ).

The solutions for ( \sin(x) = \frac{1}{2} ) are ( x = \frac{\pi}{6} + 2n\pi ) and ( x = \frac{5\pi}{6} + 2n\pi ), where ( n ) is an integer.

Combining all solutions, the general solutions for the equation ( \cos(2x)\csc(x) - 2\cos(2x) = 0 ) are ( x = \frac{\pi}{6} + 2n\pi ), ( x = \frac{5\pi}{6} + 2n\pi ), ( x = \frac{\pi}{2} + n\pi ), and ( x = \frac{3\pi}{2} + n\pi ), where ( n ) is an integer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.