The number of values of θ in the interval (-π/2, π/2) such that θ ≠ nπ/5 for n = 0, ±1, ±2 and tan θ

Q: Wait, why do we reject the negative root in the quadratic equation? Or do we?

Actually, we don't! In the quadratic $2\sin^2 2\theta + \sin 2\theta - 1 = 0$, the roots are $\sin 2\theta = \frac{1}{2}$ and $\sin 2\theta = -1$. Both are perfectly valid values for the sine function since they fall within the range $[-1, 1]$. We must explore both cases to find all possible values of $\theta$.

Q: Why can't we just use $\tan \theta = \cot 5\theta \implies \theta = \frac{\pi}{2} - 5\theta$?

That is a common trap! Trigonometric equations have periodic general solutions. The equation $\tan x = \tan y$ implies $x = k\pi + y$, not just $x = y$. By ignoring the $k\pi$ term, you would miss an infinite number of potential solutions. Always remember the general solution!

Q: How do I know if my solutions are in the excluded set?

The excluded values are multiples of $\frac{\pi}{5}$. To check, simply compare your candidate solutions (like $\frac{\pi}{12}$ or $-\frac{\pi}{4}$) with the excluded set. If a candidate can be written as $\frac{n\pi}{5}$ for some integer $n$, it's a landmine and must be discarded. In this problem, none of our candidates hit those marks, so they are all safe.

Question

The number of values of $	heta$ in the interval $(-\pi/2, \pi/2)$ such that $	heta 
eq \frac{n\pi}{5}$ for $n = 0, \pm 1, \pm 2$ and $	an 	heta = \cot 5	heta$ as well as $\sin 2	heta = \cos 4	heta$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are working within the interval $	heta \in (-\frac{\pi}{2}, \frac{\pi}{2})$. We must strictly avoid the excluded points defined by $	heta = \frac{n\pi}{5}$ for $n = 0, \pm 1, \pm 2$.

These excluded values are $\{0, \pm \frac{\pi}{5}, \pm \frac{2\pi}{5}\}$. Any candidate solution must not coincide with these values.

### Phase 1: The First Condition

We begin with the equation $	an 	heta = \cot 5	heta$. Using the co-function identity $\cot x = 	an(\frac{\pi}{2} - x)$, we rewrite the equation as:

$$	an 	heta = 	an\left(\frac{\pi}{2} - 5	hetaight)$$

Applying the general solution for the tangent function, $	an x = 	an y \implies x = k\pi + y$, we obtain:

$$	heta = k\pi + \frac{\pi}{2} - 5	heta$$

Solving for $	heta$, we find:

$$6	heta = k\pi + \frac{\pi}{2} \implies 	heta = \frac{(2k+1)\pi}{12}$$

This establishes that any valid solution must be an odd multiple of $\frac{\pi}{12}$.

### Phase 2: The Second Condition

Next, we analyze $\sin 2	heta = \cos 4	heta$. Using the double-angle identity $\cos 2A = 1 - 2\sin^2 A$ with $A = 2	heta$, the equation becomes:

$$\sin 2	heta = 1 - 2\sin^2 2	heta$$

Rearranging this into a standard quadratic form, we get:

$$2\sin^2 2	heta + \sin 2	heta - 1 = 0$$

Factoring the quadratic expression yields:

$$(2\sin 2	heta - 1)(\sin 2	heta + 1) = 0$$

This results in two distinct cases: $\sin 2	heta = \frac{1}{2}$ or $\sin 2	heta = -1$.

### Phase 3: The Intersection and Final Count

For $\sin 2	heta = \frac{1}{2}$ within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have $2	heta = \frac{\pi}{6}$ or $2	heta = \frac{5\pi}{6}$. This gives:

$$	heta = \frac{\pi}{12}, \quad 	heta = \frac{5\pi}{12}$$

For $\sin 2	heta = -1$, we have $2	heta = -\frac{\pi}{2}$, which gives:

$$	heta = -\frac{\pi}{4}$$

We now verify these candidates against our constraints. All three values ($\frac{\pi}{12}, \frac{5\pi}{12}, -\frac{\pi}{4}$) satisfy the condition $	heta = \frac{(2k+1)\pi}{12}$ and none of them are multiples of $\frac{\pi}{5}$.

The total number of valid solutions is **3**.

Animated Solution for Mathematics - Trigonometry: The number of values of $θ$ in the interval $(- π /2, π /2)$ such that $θ \neq = \frac{nπ}{5}$ for $n = 0, \pm 1, \pm 2$ and $tan θ = cot 5 θ$ as well as $sin 2 θ = cos 4 θ$ is ____.

Enter Numerical Value:

Visualized Solution

Defining the Domain

Analyzing the First Equation

General Solution for First Equation

Analyzing the Second Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Solving Case 1: $sin 2 θ = \frac{1}{2}$

Solving Case 2: $sin 2 θ = - 1$

Intersecting the Conditions

Final Constraint Check

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

Defining the Domain

Analyzing the First Equation

General Solution for First Equation

Analyzing the Second Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Solving Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ=21​

Solving Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ=−1

Intersecting the Conditions

Final Constraint Check

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Solving Case 1: $sin 2 θ = \frac{1}{2}$

Solving Case 2: $sin 2 θ = - 1$

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