Let S be the set of all α ∈ R such that the equation, cos 2x + α sin x = 2α - 7 has a solution. Then

Q: Wait, why do we reject the root $\sin x = 2$?

That is a fantastic question! Remember that the sine function, $\sin x$, is bounded. Its range is always between $-1$ and $1$ for any real number $x$. Since $2$ is outside this interval, there is no real value of $x$ that can satisfy $\sin x = 2$. Therefore, this root cannot yield a solution for $x$.

Q: Why did we choose the identity $\cos 2x = 1 - 2\sin^2 x$ instead of $2\cos^2 x - 1$?

Great observation! We look at the other term in the equation, which is $\alpha \sin x$. Since we have a $\sin x$ term, it is much easier to convert the entire equation into a quadratic in $\sin x$. If we used $2\cos^2 x - 1$, we would have a mix of $\cos^2 x$ and $\sin x$, which is harder to solve. Always aim to make the equation homogeneous in terms of a single trigonometric function.

Q: Is it possible for the discriminant to be negative?

In this specific problem, the discriminant simplifies to $(\alpha - 8)^2$. Since the square of any real number is non-negative, the discriminant is always greater than or equal to zero. This guarantees that we will always have real roots for $\sin x$ in terms of $\alpha$, which is exactly what we need to proceed with the range analysis.

Question

Let S be the set of all $\alpha \in R$ such that the equation, $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then S is equal to :

Options:

(A) [2, 6]

(B) [3, 7]

(C) R

(D) [1, 4]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation is $\cos 2x + \alpha \sin x = 2\alpha - 7$. This equation involves both a double angle and a parameter $\alpha$.

To solve this, we must unify the trigonometric terms. We utilize the identity:
$$\cos 2x = 1 - 2\sin^2 x$$

Substituting this into the original equation yields:
$$(1 - 2\sin^2 x) + \alpha \sin x = 2\alpha - 7$$

### The Quadratic Landscape

Rearranging the terms to one side, we obtain:
$$-2\sin^2 x + \alpha \sin x - 2\alpha + 8 = 0$$

Multiplying by $-1$ to simplify the leading coefficient, we get:
$$2\sin^2 x - \alpha \sin x + 2\alpha - 8 = 0$$

By substituting $t = \sin x$, where $t \in [-1, 1]$, the equation transforms into a quadratic in $t$:
$$2t^2 - \alpha t + (2\alpha - 8) = 0$$

### The Discriminant's Secret

We calculate the discriminant $D = b^2 - 4ac$ for the quadratic $2t^2 - \alpha t + (2\alpha - 8) = 0$, where $a = 2$, $b = -\alpha$, and $c = 2\alpha - 8$.

$$D = (-\alpha)^2 - 4(2)(2\alpha - 8)$$
$$D = \alpha^2 - 16\alpha + 64$$

Recognizing this as a perfect square, we have:
$$D = (\alpha - 8)^2$$

Using the quadratic formula $t = \frac{-b \pm \sqrt{D}}{2a}$, we find the roots:
$$t = \frac{\alpha \pm (\alpha - 8)}{4}$$

This yields two potential values for $\sin x$:
$$t_1 = \frac{\alpha + \alpha - 8}{4} = \frac{2\alpha - 8}{4} = \frac{\alpha - 4}{2}$$
$$t_2 = \frac{\alpha - (\alpha - 8)}{4} = \frac{8}{4} = 2$$

### The Trap and Final Calculation

We must recall the fundamental constraint of the sine function: for any real $x$, $-1 \le \sin x \le 1$. The root $t_2 = 2$ is impossible and must be rejected.

Therefore, the only valid solution arises from $t_1$:
$$-1 \le \frac{\alpha - 4}{2} \le 1$$

Multiplying the inequality by $2$:
$$-2 \le \alpha - 4 \le 2$$

Adding $4$ to all parts of the inequality, we find the range for $\alpha$:
$$2 \le \alpha \le 6$$

The set $S$ of all possible values for $\alpha$ is the closed interval **$[2, 6]$**.

Animated Solution for Mathematics - Trigonometry: Let S be the set of all $α \in R$ such that the equation, $cos 2 x + α sin x = 2 α - 7$ has a solution. Then S is equal to :

Select Answer:

Visualized Solution

Analyze the Equation

Double Angle Identity

Substitution

Quadratic Form

Discriminant Calculation

Roots of the Equation

Range of $sin x$

Condition for Solution

Solving the Inequality

Final Set $S$

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Quadratic Landscape

The Discriminant's Secret

The Trap and Final Calculation

Similar Questions

The number of solutions of $sin^{7} x + cos^{7} x = 1, x \in [0, 4 π]$ is equal to

The number of solutions of the equation $cos 2 θ cos \frac{θ}{2} + cos \frac{5 θ}{2} = 2 cos^{3} \frac{5 θ}{2}$ in $[- \frac{π}{2}, \frac{π}{2}]$ is :

The number of solutions of the equation $cos (x + \frac{π}{3}) cos (\frac{π}{3} - x) = \frac{1}{4} cos^{2} 2 x$ , $x \in [- 3 π, 3 π]$ is :

The sum of solutions of the equation $\frac{c o s x}{1 + s i n x} = ∣ tan 2 x ∣, x \in (- \frac{π}{2}, \frac{π}{2}) - {\frac{π}{4}, - \frac{π}{4}}$ is :

The number of values of $x$ in the interval $[0, 5 π]$ satisfying the equation $3 sin^{2} x - 7 sin x + 2 = 0$ is

If $0 \leq x < 2 π$ , then the number of real values of $x$ , which satisfy the equation $cos x + cos 2 x + cos 3 x + cos 4 x = 0$ is

The number of elements in the set $S = {x \in R : 2 cos (\frac{x ^{2} + x}{6}) = 4^{x} + 4^{- x}}$ is:

For $x \in (0, π)$ , the equation $sin x + 2 sin 2 x - sin 3 x = 3$ has

The real roots of the equation $cos^{7} x + sin^{4} x = 1$ in the interval $(- π, π)$ are ........., ........., and ..........

If $θ \in [- 2 π, 2 π]$ , then the number of solutions of $22 cos^{2} θ + (2 - 6) cos θ - 3 = 0$ , is equal to:

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the set of all α∈R such that the equation, cos2x+αsinx=2α−7 has a solution. Then S is equal to :