A general solution to a trigonometric equation is an expression or set of expressions that represents all possible solutions.
Because of the periodic nature of trigonometric functions, a trigonometric equation will usually have many solutions and these will occur at regular intervals. If we can find one or more basic solutions, the others can be specified by adding multiples of some fixed amount to the known solutions.
For example, the equation $\sin\theta=\frac{1}{\sqrt{3}}$sinθ=1√3 has the solution $\theta_1=\arcsin\frac{1}{\sqrt{3}}$θ1=arcsin1√3 in the first quadrant - between $0$0 and $90$90. It must also have the second quadrant solution $\theta_2=180-\arcsin\frac{1}{\sqrt{3}}$θ2=180−arcsin1√3.
We know that the sine function is periodic with period $360$360. Thus, any multiple of $360$360 added to the first or second quadrant solution will also be a solution. This means further solutions can be expressed as
$\theta_1+360n$θ1+360n and
$\theta_2+360n$θ2+360n
Guidance about finding initial solutions to trigonometric equations can be found here
Examples
SINE
Find the general solution of $\sin\theta=\frac{1}{2}$sinθ=12 in radians.
You may use $n$n to represent an integer.
We are being asked to find all the values for $\theta$θ, that make $\sin\theta=\frac{1}{2}$sinθ=12 true.
Let's just remind ourselves of the graphical solution.
trgi7
Now - let's see how to do this analytically (which just means by hand!)
Firstly we should recognise that $\sin30=\frac{1}{2}$sin30=12.
So we should use $30$30 as a reference angle to find all angles in the interval $\left[0,360\right)$[0,360) whose sine is equal to $\frac{1}{2}$12.
In the interval $\left[0,360\right)$[0,360), $\theta$θ can be both $30$30 and $150$150.
Since $\sin\theta$sinθ has a period of $360$360, then the general solution will be
$\theta_1=\alpha+360n$θ1=α+360n or $\theta_2=\beta+360n$θ2=β+360n
where $\alpha$α and $\beta$β are all the solutions in the interval $\left[0,360\right)$[0,360)
So $\alpha$α is $30$30 and $\beta$β is $150$150
Which means our general solutions are
$\theta_1=30+360n$θ1=30+360n and $\theta_2=150+360n$θ2=150+360n
Let's just check the graph again, how these general solutions map.
