Solve over the Interval tan(theta)=-(2 square root of 3)/3sin(theta) , 0< =theta< 2pi
The question provides a trigonometric equation involving a tangent function, tan(theta), and a sine function, sin(theta), and asks to solve for the values of the variable theta within a specific interval: from 0 to 2π radians. It requires finding all the angles theta that satisfy the equation when tan(theta) is set equal to the negative fraction -(2√3)/3 times sin(theta). Solving this problem involves using knowledge of trigonometric identities, relationships, and properties to isolate theta and determine its values that fulfill the equation within the given range.
$tan \left(\right. \theta\left.\right) = - \frac{2 \sqrt{3}}{3} sin \left(\right. \theta\left.\right)$,$0 \leq \theta< 2 \pi$
Answer
Solution:
Step 1: Simplify the equation.
Step 1.1: Rewrite the given equation.
Simplify $\tan(\theta) = -\frac{2\sqrt{3}}{3}\sin(\theta)$.
Step 1.1.1: Combine terms.
$\tan(\theta) = -\frac{2\sqrt{3}\sin(\theta)}{3}$.
Step 1.1.2: Rearrange the equation.
$\tan(\theta) = -\frac{2\sin(\theta)\sqrt{3}}{3}$.
Step 2: Isolate the variable.
Step 2.1: Divide by $\tan(\theta)$.
$\frac{\tan(\theta)}{\tan(\theta)} = \frac{-\frac{2\sin(\theta)\sqrt{3}}{3}}{\tan(\theta)}$.
Step 2.2: Simplify the left side.
Step 2.2.1: Cancel $\tan(\theta)$.
$1 = \frac{-\frac{2\sin(\theta)\sqrt{3}}{3}}{\tan(\theta)}$.
Step 2.3: Simplify the right side.
Step 2.3.1: Multiply by the reciprocal.
$1 = -\frac{2\sin(\theta)\sqrt{3}}{3} \cdot \frac{1}{\tan(\theta)}$.
Step 2.3.2: Express $\tan(\theta)$ as $\frac{\sin(\theta)}{\cos(\theta)}$.
$1 = -\frac{2\sin(\theta)\sqrt{3}}{3} \cdot \frac{\cos(\theta)}{\sin(\theta)}$.
Step 2.3.3: Simplify the fraction.
$1 = -\frac{2\sqrt{3}}{3}\cos(\theta)$.
Step 3: Solve for $\cos(\theta)$.
Step 3.1: Isolate $\cos(\theta)$.
$\cos(\theta) = -\frac{3}{2\sqrt{3}}$.
Step 4: Find the angles.
Step 4.1: Take the inverse cosine.
$\theta = \arccos\left(-\frac{\sqrt{3}}{2}\right)$.
Step 4.2: Determine the angles in the specified interval.
$\theta = \frac{5\pi}{6}, \frac{7\pi}{6}$.
Knowledge Notes:
Trigonometric identities: The tangent of an angle is the ratio of the sine to the cosine of that angle, $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.
Inverse trigonometric functions: To find an angle given a trigonometric value, we use the inverse trigonometric functions, such as $\arccos(x)$, which returns the angle whose cosine is $x$.
The cosine function is negative in the second and third quadrants of the unit circle.
Periodicity of trigonometric functions: The trigonometric functions repeat their values in regular intervals called periods. For the cosine function, the period is $2\pi$ radians.
Interval notation: When solving trigonometric equations, it is important to consider the interval in which the solution is sought. For example, $0 \leq \theta < 2\pi$ means that the angle $\theta$ must be between $0$ and $2\pi$, including $0$ but not $2\pi$.
Simplifying expressions: When simplifying expressions, it is often useful to cancel common factors, combine like terms, and use algebraic properties to rewrite the expressions in a simpler form.
