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1+1/sin(alpha) =4 the range is for alpha is from 0 to 360 find alpha???

I got the first angle as 19.5, however, the marking scheme stated that there is another angle(160.5) can you please explain how they got the second angle

Answer 1

Caroline Adams

See below.

We have:

#(sin alpha + 1)/sinalpha = 4#

#sinalpha + 1 = 4sinalpha #

#1 = 3sinalpha#

#sinalpha = 1/3#

#alpha = 19.5 and ?#

We know that sine is positive in the first and second quadrants, thus the reference angle for the second value of alpha is going to be #19.5˚#, or #alpha_2 = 180˚ - 19.5˚ = 160.5˚#.

As required.

Hopefully this helps!

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Answer 2

Charles Autterson

To find the values of ( \alpha ) that satisfy the equation ( 1 + \frac{1}{\sin(\alpha)} = 4 ) within the range ( 0^\circ ) to ( 360^\circ ), follow these steps:

Subtract 1 from both sides of the equation to isolate ( \frac{1}{\sin(\alpha)} ).
Find the reciprocal of both sides to solve for ( \sin(\alpha) ).
Solve for ( \alpha ) by taking the arcsine of both sides.
Consider the range of ( \alpha ) to be from ( 0^\circ ) to ( 360^\circ ).
Evaluate the arcsine expression for ( \alpha ) within this range.

The solutions for ( \alpha ) will be the values within the specified range that satisfy the equation.

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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.