# quite urgent..exam in 2 days!!

• May 13th 2008, 03:07 AM
Helpppp
quite urgent..exam in 2 days!!
just doing some past papers and there is a type of question that I can't understand..ever! (Headbang)

for example:
solve the equation: sin(θ - 30°) = 0.7 in the interval 0°≤ θ ≤ 360°
solve the equation: cosx=0.3 in the interval 0 ≤ x ≤ 2π
thank you!
(Hi)
• May 13th 2008, 03:21 AM
Gusbob
Basically, you are asked to solve for the variable. For example, your first question:

$sin(\theta-30^\circ) = 0.7$
$\theta - 30^\circ = arcsin(0.7)$
$\theta = arcsin(0.7)+30^\circ$

Now punch that through your calculator. arcsin is more often this button: $sin^{-1}$

Remember that there are two quadrants in the unit circle where sin is positive, so in a complete circle (360 degrees), there are two solutions to this problem.

Now do the second one in the same way
• May 13th 2008, 03:24 AM
Helpppp
how do I relate this to the interval?
I know it's something to do with the graphs for sin and cos but how do I work out the answers for the interval given?
thanks
(Rofl)
• May 13th 2008, 03:36 AM
Gusbob
One Method:
Use the unit circle. One revolution of the circle is 360 degrees (or 2pi). Sine is positive in the 1st and 2nd quadrants. So the solutions to consider are is $\theta$ and $180°-\theta$.
__________________________________________________ _____________
If you are unfamiliar with the unit circle, just use the general solutions for the sine:
Having found the basic value of $\theta$ (which is $arcsin(0.7)+30^\circ$), substitute it into the general equation for the sin.

$x = 180^\circ n + (-1)^n (\theta)$
where x is a solution, and n is an integer larger or equal to zero (0,1,2,3...)

Substitute n =0 to get the first solution, n= 1 for the second, n= 2 for the third and so on...

You'll find that after n = 1, the angles will become larger than 360°, and thus not within your parameters. That leavs you with two solutions.