Solve the following equations if 0<x<2pie (those are 'less than or equal to') answer to 2 decimal places.
a) square root of 2 cos(x = pie/2) + 1 = 0
b) sin(1/2)(x-2pie/9) = 0.6
i would be greatful to get step by step answers please
Tom
Solve the following equations if 0<x<2pie (those are 'less than or equal to') answer to 2 decimal places.
a) square root of 2 cos(x = pie/2) + 1 = 0
b) sin(1/2)(x-2pie/9) = 0.6
i would be greatful to get step by step answers please
Tom
I'd love to help but have problems understanding things on both questions.
For a) , what does 2cos(x = pi/2) mean? Is that supposed to be a minus sign?
For b) , is that supposed to be the sine of one half time the next expression, kind of like (sin .5) times (x-[2pi]/p)?
Also, for A, does the square root apply to the whole left hand side or only part of it. Use those parantheses!
B) I don't have a good calculator handy, so I'll just solve in terms of what you wrote and you can make it into a decimal answer. This is really just like any algebra problem you've gotten before. Isolate x.
sin(1/2) * (x-[2pi]/9) = .6
x-[2pi]/9 = (.6/sin[1/2])
x = (.6/sin[1/2]) + [2pi]9
From here, it's just making sure you can enter it properly into your calculator to get the right decimal approximation.
This could be done using a inverse trig method, but a much easier and in my opinion a much more elegant way uses a common but wonderful observation.
How are sine and cosine directly related? Well here's an identity that will prove useful.
cos(a) = sin(pi/2 - a)
So our problem is:
sqrt{2}cos(x+pi/2) + 1 = 0
Now, for this case the "a" is the whole expression "x+pi/2". So we can rewrite this as...
sqrt{2}sin(pi/2-a) + 1 = 0
And plugging back in our a...
sqrt{2}sin(pi/2 - [x+pi/2]) + 1 = 0
Simplify...
sqrt{2}sin(-x) + 1 = 0
sin(-x) = -1 / sqrt{2} or -sqrt{2}/2
From here you can use your calculator, but it's not necessary. This is a nice angle to work with.