Math / Trigonometry

QuestionSolve the equation. $2 \sin ^{2} \theta-3 \sin \theta+1=0$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $\theta$ that make this tricky equation using sines true! Watch out! Remember that when solving trigonometric equations, there can be multiple solutions, so we need to be extra careful not to miss any!
Also, watch out for those pesky domain restrictions!

STEP 2

1. Rewrite the equation
2. Solve the quadratic equation
3. Find the angles

STEP 3

Let's make this equation look a little friendlier!
We can substitute $x = \sin\theta$ .
This gives us a nice quadratic equation: $2x^2 - 3x + 1 = 0$

STEP 4

Now, we can factor this quadratic equation.
We're looking for two numbers that multiply to $2 \cdot 1 = \textbf{2}$ and add up to $\textbf{-3}$ .
Those numbers are $\textbf{-2}$ and $\textbf{-1}$ ! $2x^2 - 2x - x + 1 = 0$

STEP 5

Let's factor by grouping: $2x(x - 1) - 1(x - 1) = 0$ $(2x - 1)(x - 1) = 0$

STEP 6

So, either $2x - 1 = 0$ or $x - 1 = 0$ .
This gives us two possible solutions for $x$ : $2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$ $x - 1 = 0 \implies x = 1$

STEP 7

Remember, $x = \sin\theta$ .
So we have two equations to solve for $\theta$ : $\sin\theta = \frac{1}{2}$ $\sin\theta = 1$

STEP 8

For $\sin\theta = \frac{1}{2}$ , we know that $\theta$ can be $\frac{\pi}{6}$ or $\frac{5\pi}{6}$ in the interval $[0, 2\pi)$ .
Remember, sine is positive in the first and second quadrants!

STEP 9

For $\sin\theta = 1$ , we know that $\theta = \frac{\pi}{2}$ in the interval $[0, 2\pi)$ .
Sine is one at the top of the unit circle!

STEP 10

The solutions for $\theta$ in the interval $[0, 2\pi)$ are $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ , and $\frac{\pi}{2}$ .
Don't forget, if the problem doesn't specify an interval, we would generally write the general solutions as: $\theta = \frac{\pi}{6} + 2n\pi$ , $\theta = \frac{5\pi}{6} + 2n\pi$ , and $\theta = \frac{\pi}{2} + 2n\pi$ , where $n$ is any integer.

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