Question

Cos²x/4-sin²x/4>-0.5

Answer 1

Cos²x заменяем на 1-sin²x получается -2sin²x=-0,5  
делим на -2 получается sin²x=1/4 
sinx=1/2
получается x=(-1)^k*π/6+πk

Answer 2

To solve the inequality $\frac{\cos^2x}{4}-\frac{\sin^2x}{4} > -0.5$, we can simplify and manipulate the equation step-by-step. Here's the process:

Step 1: Combine the fractions on the left-hand side: $\frac{\cos^2x - \sin^2x}{4} > -0.5$.

Step 2: Use the trigonometric identity $\cos^2x - \sin^2x = \cos(2x)$: $\frac{\cos(2x)}{4} > -0.5$.

Step 3: Multiply both sides by 4 to eliminate the fraction (note: since we're multiplying by a positive number, the inequality sign doesn't change): $\cos(2x) > -2$.

Step 4: Solve for $x$ by taking the inverse cosine (arccos) of both sides. Remember that the inverse cosine function has a range of $0 \leq \text{arccos}(x) \leq \pi$: $2x > \text{arccos}(-2)$.

Step 5: Divide both sides by 2 to get the final answer for $x$: $x > \frac{\text{arccos}(-2)}{2}$.

Now, we need to be cautious with the range of the inverse cosine function. The value of $\text{arccos}(x)$ is defined for $x$ in the range $-1 \leq x \leq 1$. However, $-2$ is outside this range, which means there are no real values of $x$ that satisfy the given inequality. In other words, the inequality has no solutions in the real domain.

Keep in mind that the range of $\text{arccos}(x)$ is $[0, \pi]$, so the fraction $\frac{\text{arccos}(-2)}{2}$ is not a real number. Therefore, the inequality is not valid for any real $x$.