# Thread: quite urgent..exam in 2 days!!

1. ## quite urgent..exam in 2 days!!

just doing some past papers and there is a type of question that I can't understand..ever! 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
can anyone help me please??
thank you! 2. Basically, you are asked to solve for the variable. For example, your first question:

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

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

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

3. 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 4. 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 $\displaystyle \theta$ and $\displaystyle 180°-\theta$.
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If you are unfamiliar with the unit circle, just use the general solutions for the sine:
Having found the basic value of $\displaystyle \theta$ (which is $\displaystyle arcsin(0.7)+30^\circ$), substitute it into the general equation for the sin.

$\displaystyle x = 180^\circ n + (-1)^n (\theta)$
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.

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