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

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- April 11th 2007, 12:43 AMT_o_Msome trignometry simplification
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 - April 11th 2007, 01:28 AMJameson
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)? - April 11th 2007, 01:39 AMT_o_M
sorry

for a) square root of 2 cos(x + pi/2) + 1 = 0

for b) yes it is the sine of one half time the next expression, kind of like (sin .5) times (x-[2pi]/9) = 0.6 - April 11th 2007, 02:00 AMJameson
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. - April 11th 2007, 02:03 AMT_o_M
for a) square root ONLY applies to the 2 at the front

- April 11th 2007, 02:21 AMJameson
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.