General Solution of Trigonometric Equations

$$\sqrt{3}\cos2\theta+8\cos\theta+3\sqrt{3}=0$$

We know that $$\cos2\theta=2cos^2\theta-1$$

$$\sqrt{3}\left(2\cos^2\theta-1\right)+8\cos\theta+3\sqrt{3}=0$$ 

$$2\sqrt{3}\cos^2\theta-\sqrt{3}+8\cos\theta+3\sqrt{3}=0$$

$$2\sqrt{3}\cos^2\theta+8\cos\theta+2\sqrt{3}=0$$

$$2\sqrt{3}\cos^2\theta+6\cos\theta+2\cos\theta+2\sqrt{3}=0$$

$$2\sqrt{3}\cos\theta(\cos\theta+\sqrt{3})+2(\cos\theta+\sqrt{3})=0$$

$$(2\sqrt{3}\cos\theta+2)(\cos\theta+\sqrt{3})=0$$

Thus, there are two solutions possible -> $$\cos\theta=-\dfrac{1}{\sqrt{3}}$$ and $$\cos\theta=-\sqrt{3}$$

Since, the value of $$\cos\theta\epsilon[-1,1]$$, therefore, $$\cos\theta=-\sqrt{3}$$ is never possible. 

Now, $$\cos\theta$$ is negative in the second and the third quadrant, and in each cycle of $$\pi$$, $$\cos\theta$$ takes all values in $$[-1,1]$$. Thus, in every cycle of $$\pi$$, there will be one solution for $$\cos\theta=-\dfrac{1}{\sqrt{3}}$$.

Thus, from $$[-3\pi,2\pi]$$, there will be 5 solutions for this equation.

Step 1: Grouping and Converting to Sine and Cosine

We start by bringing the single tangent term to the denominator on the Left Hand Side to group our variables logically.

$$\frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ) \tan(x - 50^\circ)$$

Next, we break everything down into sines and cosines using the fundamental identity $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$.

For the Left Hand Side, we have a fraction divided by a fraction. When dividing by a fraction, we multiply by its reciprocal.

$$\text{Left Side} = \frac{\frac{\sin(x+100^\circ)}{\cos(x+100^\circ)}}{\frac{\sin x}{\cos x}}$$
$$\text{Left Side} = \frac{\sin(x+100^\circ)}{\cos(x+100^\circ)} \times \frac{\cos x}{\sin x} = \frac{\sin(x+100^\circ) \cos x}{\cos(x+100^\circ) \sin x}$$

For the Right Hand Side, we simply multiply two tangent fractions straight across the numerators and denominators.

$$\text{Right Side} = \left( \frac{\sin(x+50^\circ)}{\cos(x+50^\circ)} \right) \left( \frac{\sin(x-50^\circ)}{\cos(x-50^\circ)} \right) = \frac{\sin(x+50^\circ) \sin(x-50^\circ)}{\cos(x+50^\circ) \cos(x-50^\circ)}$$

Equating our newly expanded sides gives us our working foundation:

$$\frac{\sin(x+100^\circ) \cos x}{\cos(x+100^\circ) \sin x} = \frac{\sin(x+50^\circ) \sin(x-50^\circ)}{\cos(x+50^\circ) \cos(x-50^\circ)}$$

Step 2: Applying Componendo and Dividendo

If $$\frac{a}{b} = \frac{c}{d}$$, we can instantly rewrite it as $$\frac{a+b}{a-b} = \frac{c+d}{c-d}$$. This allows us to rapidly build standard trigonometric identities.

Processing the Left Hand Side:

The new numerator becomes $$a+b$$:

$$\sin(x+100^\circ) \cos x + \cos(x+100^\circ) \sin x = \sin(x+100^\circ + x) = \sin(2x + 100^\circ)$$

The new denominator becomes $$a-b$$:

$$\sin(x+100^\circ) \cos x - \cos(x+100^\circ) \sin x = \sin(x+100^\circ - x) = \sin 100^\circ$$

Processing the Right Hand Side:

The new numerator becomes $$c+d$$:

$$\sin(x+50^\circ) \sin(x-50^\circ) + \cos(x+50^\circ) \cos(x-50^\circ) = \cos((x+50^\circ) - (x-50^\circ)) = \cos 100^\circ$$

The new denominator becomes $$c-d$$:

$$\sin(x+50^\circ) \sin(x-50^\circ) - \cos(x+50^\circ) \cos(x-50^\circ)$$
$$= -[ \cos(x+50^\circ) \cos(x-50^\circ) - \sin(x+50^\circ) \sin(x-50^\circ) ] = -\cos(x+50^\circ + x-50^\circ) = -\cos 2x$$

Putting the new simplified fractions together, our massive equation shrinks down to:

$$\frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x}$$

Step 3: Cross Multiplication and Double Angle Identities

Cross multiply to remove the fractions completely:

$$-\sin(2x + 100^\circ) \cos 2x = \sin 100^\circ \cos 100^\circ$$

Multiply both sides of the equation by 2. This perfectly sets up the product to sum formula $$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$ on the left, and the double angle formula $$2\sin \theta \cos \theta = \sin 2\theta$$ on the right.

$$-2\sin(2x + 100^\circ) \cos 2x = 2\sin 100^\circ \cos 100^\circ$$
$$-[ \sin(2x + 100^\circ + 2x) + \sin(2x + 100^\circ - 2x) ] = \sin 200^\circ$$
$$-[ \sin(4x + 100^\circ) + \sin 100^\circ ] = \sin 200^\circ$$

Move $$\sin 100^\circ$$ to the other side to isolate our $$x$$ variable:

$$-\sin(4x + 100^\circ) = \sin 200^\circ + \sin 100^\circ$$

Step 4: Sum to Product Conversion

We simplify the right side using the formula $$\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$:

$$\sin 200^\circ + \sin 100^\circ = 2\sin\left(\frac{300^\circ}{2}\right)\cos\left(\frac{100^\circ}{2}\right) = 2\sin 150^\circ \cos 50^\circ$$

Because $$\sin 150^\circ = \frac{1}{2}$$, this entire term collapses cleanly into just $$\cos 50^\circ$$.

$$-\sin(4x + 100^\circ) = \cos 50^\circ$$
$$\sin(4x + 100^\circ) = -\cos 50^\circ$$

Step 5: Solving for the Final Angles

To solve the equation, we need matching trigonometric functions. We must convert the negative cosine into a sine function.

$$-\cos 50^\circ = -\sin 40^\circ = \sin(-40^\circ)$$

This gives us our final trigonometric equation to solve:

$$\sin(4x + 100^\circ) = \sin(-40^\circ)$$

We plug this into the standard general solution formula $$\theta = n(180^\circ) + (-1)^n \alpha$$:

$$4x + 100^\circ = n(180^\circ) + (-1)^n (-40^\circ)$$

Now we test consecutive integer values for $$n$$ to discover the specific solutions that fit into the given interval between $$0^\circ$$ and $$180^\circ$$.

• For n = 1:

  $$4x + 100^\circ = 180^\circ - (-40^\circ) = 220^\circ$$
  $$4x = 120^\circ \implies x = 30^\circ$$

• For n = 2:

  $$4x + 100^\circ = 360^\circ + (-40^\circ) = 320^\circ$$
  $$4x = 220^\circ \implies x = 55^\circ$$

• For n = 3:

  $$4x + 100^\circ = 540^\circ - (-40^\circ) = 580^\circ$$
  $$4x = 480^\circ \implies x = 120^\circ$$

• For n = 4:

  $$4x + 100^\circ = 720^\circ + (-40^\circ) = 680^\circ$$
  $$4x = 580^\circ \implies x = 145^\circ$$

If we try $$n = 0$$, we get $$x = -35^\circ$$, and if we try $$n = 5$$, we get $$x = 210^\circ$$. Both of these fall strictly outside the required domain boundaries.

The valid angles are exactly $$30^\circ$$, $$55^\circ$$, $$120^\circ$$, and $$145^\circ$$.

The final number of elements in the set is 4.

We wish to find the number of solutions of $$\sin^2 x + (2+2x-x^2)\sin x - 3(x-1)^2 = 0$$ for $$-\pi \leq x \leq \pi$$.

Letting $$s = \sin x$$ and observing that $$2+2x-x^2 = 3-(x-1)^2$$ transforms the equation into $$s^2 + \bigl(3-(x-1)^2\bigr)s - 3(x-1)^2 = 0,$$ which can be rewritten as $$s^2 + 3s - (x-1)^2(s+3) = 0$$ and thus factors to $$(s+3)(s-(x-1)^2) = 0.$$

It follows that $$\sin x + 3 = 0$$, giving $$\sin x = -3$$, which is impossible since $$|\sin x|\leq1$$. Therefore we must have $$\sin x = (x-1)^2$$, requiring $$0 \leq (x-1)^2 \leq 1$$ and hence $$x \in [0,2].$$

On the interval $$[0,2]$$, comparing $$y = \sin x$$ and $$y = (x-1)^2$$ shows that at $$x = 0$$, $$\sin0 = 0$$ is below $$(0-1)^2 = 1$$; at $$x = 1$$, $$\sin1 \approx 0.841$$ is above $$(1-1)^2 = 0$$; and at $$x = 2$$, $$\sin2 \approx 0.909$$ is below $$(2-1)^2 = 1$$. Thus the two curves intersect twice, once in $$(0,1)$$ and once in $$(1,2)$$.

Therefore there are 2 solutions, and the correct answer is $$\boxed{2}$$.