Si 0° ≤ θ < 360°, ¿cuáles son las soluciones para tan²θ + 6 = 6?
Understand the Problem
La pregunta pide encontrar las soluciones para la ecuación trigonométrica ( \tan^2(\theta) + 6 = 6 ) dentro del rango ( 0° \leq \theta < 360° ). Primero, simplificaremos la ecuación y luego encontraremos los valores de ( \theta ) que satisfacen la ecuación resultante dentro del rango especificado.
Answer
$\theta = 0^\circ, 180^\circ$
Answer for screen readers
The solutions for $\tan^2(\theta) + 6 = 6$ in the range $0^\circ \leq \theta < 360^\circ$ are: $$ \theta = 0^\circ, 180^\circ $$
Steps to Solve
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Simplify the equation Subtract 6 from both sides of the equation. $$ \tan^2(\theta) + 6 = 6 $$ $$ \tan^2(\theta) = 0 $$
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Solve for $\tan(\theta)$ Take the square root of both sides to find $\tan(\theta)$. $$ \sqrt{\tan^2(\theta)} = \sqrt{0} $$ $$ \tan(\theta) = 0 $$
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Find the angles $\theta$ where $\tan(\theta) = 0$ within the given range The tangent function is zero at multiples of $180^\circ$. Within the range $0^\circ \leq \theta < 360^\circ$, the angles are: $$ \theta = 0^\circ, 180^\circ $$
The solutions for $\tan^2(\theta) + 6 = 6$ in the range $0^\circ \leq \theta < 360^\circ$ are: $$ \theta = 0^\circ, 180^\circ $$
More Information
The tangent function is zero at angles where the sine function is zero. This occurs at $0^\circ$, $180^\circ$, $360^\circ$, etc. Since the range is $0^\circ \leq \theta < 360^\circ$, $360^\circ$ is not included in the solutions.
Tips
A common mistake is forgetting to include $180^{\circ}$ in the solution set, as $\tan(180^{\circ}) = 0$. Another common mistake is including $360^{\circ}$ in the solution set, however the problem specifies $\theta < 360^{\circ}$.
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