Question

2 - 3 sin(3pi/2 + x) + cos^2(x/2) = sin^2(x/2)

Answer 1

2-3sin(3π/2+x)+cos²(x/2)-sin²(x/2)=0
2+3sinx+cos²(x/2)-sin²(x/2)=0
2sin²(x/2)+2cos²(x/2)+6sin(x/2)cos(x/2)+cos²(x/2)-sin²(x/2)=0
sin²(x/2)+6sin(x/2)cos(x/2)+3cos²(x/2)=0/cos²(x/2)
tg²(x/2)+6tg(x/2)+3=0
tg(x/2)=a
a2+6a+3=0
D=36-12=24
a1=(-6-2√6)/2=-3-√6⇒tg(x/2)=-3-√6⇒x/2=-arctg(3+√6) +πn⇒x=-1arctg(3+√6) +2πn
a2=-3+√6⇒tg(x/2)=√6-3⇒x/2=arctg(√6-3)+πn⇒x=2arctg(√6-3)+2πn

Answer 2

To solve the equation $2 - 3 \sin\left(\frac{3\pi}{2} + x\right) + \cos^2\left(\frac{x}{2}\right) = \sin^2\left(\frac{x}{2}\right)$, we'll use trigonometric identities and algebraic manipulation.

1. Use Angle Sum and Difference Identities:

   • $\sin\left(\frac{3\pi}{2} + x\right) = \sin\frac{3\pi}{2} \cos(x) + \cos\frac{3\pi}{2} \sin(x) = -\cos(x)$.
   • $\cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos(x)}{2}$.
   • $\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{2}$.

2. Substitute these Identities into the Equation:

   $$2 - 3(-\cos(x)) + \frac{1 + \cos(x)}{2} = \frac{1 - \cos(x)}{2}$$

3. Simplify the Equation:

   $$2 + 3\cos(x) + \frac{1 + \cos(x)}{2} = \frac{1 - \cos(x)}{2}$$

4. Get rid of Fractions: Multiply every term by 2 to eliminate the fractions:

   $$4 + 6\cos(x) + 1 + \cos(x) = 1 - \cos(x)$$

5. Combine Like Terms:

   $$7\cos(x) + 5 = -\cos(x)$$

6. Move Terms Around:

   $$8\cos(x) = -5$$

7. Isolate $\cos(x)$:

   $$\cos(x) = -\frac{5}{8}$$

8. Find $x$: Use the inverse cosine function:

   $$x = \cos^{-1}\left(-\frac{5}{8}\right)$$

   Note: This solution is valid if $x$ is allowed to take any real value. If $x$ is constrained to a specific interval, the solution might change.

Keep in mind that the solution might not be in a simple closed form, and you might need to use a calculator to get a numerical approximation.