Question

16sin2x+sin2x−12cos2x=1

Answer 1

Ответ:

y=16sin2x+sin2x-12cos2x

y=1

xє (0, -12)

Объяснение:

Answer 2

To solve the trigonometric equation $16\sin^2x + \sin^2x - 12\cos^2x = 1$, we can use trigonometric identities to simplify it and solve for $x$. The fundamental trigonometric identity that relates the sine and cosine functions is:

$\sin^2x + \cos^2x = 1$

Let's use this identity to simplify the equation step by step:

Given equation: $16\sin^2x + \sin^2x - 12\cos^2x = 1$

Combine like terms: $17\sin^2x - 12\cos^2x = 1$

Now, we can express $\cos^2x$ in terms of $\sin^2x$ using the identity $\cos^2x = 1 - \sin^2x$:

$17\sin^2x - 12(1 - \sin^2x) = 1$

Distribute the 12: $17\sin^2x - 12 + 12\sin^2x = 1$

Combine terms: $29\sin^2x - 12 = 1$

Now, isolate $\sin^2x$: $29\sin^2x = 13$

Divide by 29: $\sin^2x = \frac{13}{29}$

Take the square root of both sides: $\sin x = \pm \sqrt{\frac{13}{29}}$

Since $\sin x$ is positive in the first and second quadrants, we only consider the positive square root:

$\sin x = \sqrt{\frac{13}{29}}$

Now, to find the actual value of $x$, take the arcsine (inverse sine) of both sides:

$x = \arcsin\left(\sqrt{\frac{13}{29}}\right)$

This will give you the value of $x$ in radians. If you want the value in degrees, you can convert it using the fact that $180^\circ = \pi$ radians.