## Saturday, February 13, 2016

### High School Math Solutions – Trigonometry Calculator, Trig Equations

In a previous post, we learned about trig evaluation. It is important that topic is mastered before continuing on. In this blog post, we will talk about trig equations, which involve trig evaluation. Solving trig equations are similar to solving algebraic equations. We are given an unknown variable, which we have to solve for. In order to solve for the unknown, we are able to factor an expression, multiply or divide, square or take the square root, and substitute identities. Solving trig equations can be a little tricky because there is no standard procedure to solving trig equations.

Let’s see some tips to help us solve trig equations  . . .

1. If there is more than one trig function, rewrite the equation in terms of only one trig function. This will require using trig identities and, or algebraic manipulation.
2. Use factoring if you see an expression in quadratic form.
3. Memorize the unit circle
4. Solving trig equations is all about trial and error. If you get it wrong the first time, try solving it a different way.

Solving trig equations aren’t very bad. Let’s see how to solve one step by step.

2\sin ^2\left(x\right)+3=7\sin \left(x\right),\:0\le \:x\le \:2\pi

1. We can see that the trigonometric expression is written in quadratic form. Let’s move everything to one side of the equation.
\sin^2(x)-7\sin(x)+3=0
2. To make the expression look nicer, we will let u=sin(x).
2u^2-7u+3=0
3. Now, we see much easier how to factor the function. So we will solve for u.
(2u-1)(u-3)=0
u=\frac{1}{2}\:or\:u=3
4. We’ll substitute back sin(x)
\sin(x)=\frac{1}{2}\:or\:\sin(x)=3
5. Now we have to check the constraints
\sin(x)=\frac{1}{2},\:0\le x\le 2\pi
\sin(x)=3,\:0\le x\le 2\pi
\sin(x)=\frac{1}{2},\:0\le x\le2\pi,\:when\:x=\frac{\pi}{6}\:or\:x=\frac{5\pi}{6}
\sin(x)\ne3, so this has no solutions

6. Put all the solutions together
x=\frac{\pi}{6}\:or\:x=\frac{5\pi}{6}