Question

6sin^2(pi-x)-1,5cos(pi/2-2x)-cos^2x=1

Answer 1

6sin²(π - x) - 1,5cos(π/2 - 2x) - cos²x = 1

6sin²x - 1,5sin2x - cos²x = 1

6sin²x - 1,5•2•sinx•cosx - cos²x = sin²x + cos²x

5sin²x - 3•sinx•cosx - 2cos²x = 0

Разделим обе части cos²x ≠ 0

5tg²x - 3tgx - 2 = 0

Замена tgx = a, a ∈ R

5a² - 3a - 2 = 0

D = (-3)² - 4•5•(-2) = 9 + 40 = 49

a₁ = (3 - 7)/10 = - 4/10 = - 2/5  ⇒ tgx = - 2/5 ⇔ x = - arctg(2/5) + πn, n ∈ Z

a₂ = (3 + 7)/10 = 10/10 = 1  ⇒ tgx = 1 ⇔ x = π/4 + πk, k ∈ Z

ОТВЕТ: - arctg(2/5) + πn ; π/4 + πk , n , k ∈ Z

Answer 2

To solve the equation $6\sin^2(\pi - x) - 1 + 5\cos\left(\frac{\pi}{2} - 2x\right) - \cos^2x = 1$, we'll simplify the trigonometric expressions step by step and then find the solutions for $x$.

Step 1: Use Trigonometric Identities We can use the following trigonometric identities:

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

Step 2: Rewrite the equation using the above identities The equation becomes: $6\sin^2 x - 1 + 5\sin 2x - \cos^2x = 1$

Step 3: Combine like terms $6\sin^2 x - \cos^2x + 5\sin 2x = 2$

Step 4: Use double angle identity for sine $6\sin^2x - \cos^2x + 10\sin x\cos x = 2$

Step 5: Use the Pythagorean identity $\cos^2x = 1 - \sin^2x$ $6\sin^2x - (1 - \sin^2x) + 10\sin x\cos x = 2$

Step 6: Combine terms again $7\sin^2x + 10\sin x\cos x - 1 = 2$

Step 7: Rearrange the equation $7\sin^2x + 10\sin x\cos x - 3 = 0$

Now, we have a quadratic-like equation in terms of $\sin x$. Let's denote $\sin x$ as $t$:

$7t^2 + 10t - 3 = 0$

Step 8: Solve for $t$ We can use the quadratic formula to solve for $t$: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 7$, $b = 10$, and $c = -3$.

$t = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 7 \cdot (-3)}}{2 \cdot 7}$

$t = \frac{-10 \pm \sqrt{100 + 84}}{14}$

$t = \frac{-10 \pm \sqrt{184}}{14}$

Step 9: Simplify further $t = \frac{-10 \pm 2\sqrt{46}}{14}$

$t = \frac{-5 \pm \sqrt{46}}{7}$

So we have two possible values for $\sin x$:

1. $\sin x = \frac{-5 + \sqrt{46}}{7}$
2. $\sin x = \frac{-5 - \sqrt{46}}{7}$