Solving the Trigonometric Equation for θ

The question asks us to find the value of $\theta$ within the range $0^\circ \le \theta \le 90^\circ$ for which the given trigonometric equation $\sin^2 \theta - 3 \sin \theta + 2 = 0$ holds true.

The given equation is:

$$ \sin^2 \theta - 3 \sin \theta + 2 = 0 $$

This equation looks like a quadratic equation if we consider $\sin \theta$ as a variable. Let $x = \sin \theta$. Substituting $x$ into the equation, we get:

$$ x^2 - 3x + 2 = 0 $$

Now, we need to solve this quadratic equation for $x$. We can factor the quadratic expression:

We are looking for two numbers that multiply to $2$ and add up to $-3$. These numbers are $-1$ and $-2$.

$$ x^2 - 1x - 2x + 2 = 0 $$

$$ x(x - 1) - 2(x - 1) = 0 $$

$$ (x - 1)(x - 2) = 0 $$

This gives us two possible solutions for $x$:

• $x - 1 = 0 \implies x = 1$
• $x - 2 = 0 \implies x = 2$

Now, we substitute back $x = \sin \theta$:

• $\sin \theta = 1$
• $\sin \theta = 2$

Analyzing the Values of sin θ

We know that the value of the sine function, $\sin \theta$, can only range between $-1$ and $1$ for any real angle $\theta$. That is, $-1 \le \sin \theta \le 1$.

Let's examine our solutions for $\sin \theta$:

• Case 1: $\sin \theta = 1$

  This value is within the valid range $[-1, 1]$. We need to find the angle $\theta$ between $0^\circ$ and $90^\circ$ for which $\sin \theta = 1$. Looking at standard trigonometric values, we know that $\sin 90^\circ = 1$. Since the range specified is $0^\circ \le \theta \le 90^\circ$, $\theta = 90^\circ$ is a valid solution.

• Case 2: $\sin \theta = 2$

  This value is outside the valid range $[-1, 1]$ for $\sin \theta$. There is no real angle $\theta$ for which $\sin \theta$ is equal to $2$. Therefore, $\sin \theta = 2$ does not give a valid solution for $\theta$.

Based on our analysis, the only valid solution for $\sin \theta$ is $1$, which corresponds to $\theta = 90^\circ$ in the given range $0^\circ \le \theta \le 90^\circ$.

Conclusion for the Value of θ

The value of $\theta$ that satisfies the equation $\sin^2 \theta - 3 \sin \theta + 2 = 0$ in the range $0^\circ \le \theta \le 90^\circ$ is $\theta = 90^\circ$.

Revision Table: Common Trigonometric Values (0° to 90°)

Additional Information: Quadratic Equations and Trigonometry

This problem combined two important concepts: solving quadratic equations and understanding trigonometric functions.

• Quadratic Equations: An equation of the form $ax^2 + bx + c = 0$, where $a \ne 0$, is a quadratic equation. They can often be solved by factoring, using the quadratic formula ($x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$), or completing the square. In this problem, the variable was $\sin \theta$ instead of a simple $x$.
• Range of Sine Function: The sine function, $\sin \theta$, takes values between $-1$ and $1$ for any real angle $\theta$. Graphically, the sine wave oscillates between $y=-1$ and $y=1$. Understanding this range is crucial when solving trigonometric equations, as it helps filter out invalid solutions obtained from algebraic steps.
• Solving Trigonometric Equations: These often require algebraic manipulation to isolate the trigonometric function, followed by finding the angles that satisfy the resulting equation within a specified domain (like $0^\circ \le \theta \le 90^\circ$). Knowledge of inverse trigonometric functions and unit circle or standard angle values is helpful.

Always remember to check if the solutions obtained for the trigonometric function (like $\sin \theta$ or $\cos \theta$) are within their valid range before finding the angle $\theta$.