Question

3cos^2x-sin^2+2sinx*cosx=0

Answer 1

3cos²x-sin²x+2sinxcosx=0/cos²x≠0
tg²x-2tgx-3=0
tgx=a
a²-2a-3=0
a1+a2=2 U a1*a2=-3
a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈Z
a2=3⇒tgx=3⇒x=arctg3+πk,k∈Z

Answer 2

To solve the trigonometric equation $3\cos^2(x) - \sin^2(x) + 2\sin(x)\cos(x) = 0$, we can use trigonometric identities to simplify it.

Start by using the double angle identity $\sin(2x) = 2\sin(x)\cos(x)$. Substitute this identity into the equation:

$3\cos^2(x) - \sin^2(x) + \sin(2x) = 0.$

Now, we have an equation involving both $\cos(x)$ and $\sin(x)$. We can use another trigonometric identity, the Pythagorean identity, which states that $\sin^2(x) + \cos^2(x) = 1$. We can rewrite $\sin^2(x)$ as $1 - \cos^2(x)$ and substitute it into the equation:

$3\cos^2(x) - (1 - \cos^2(x)) + \sin(2x) = 0.$

Now, simplify the equation:

$3\cos^2(x) - 1 + \cos^2(x) + \sin(2x) = 0.$

Combine like terms:

$4\cos^2(x) - 1 + \sin(2x) = 0.$

Now, we have a quadratic equation in terms of $\cos(x)$. Let's solve it:

$4\cos^2(x) - 1 = -\sin(2x).$

$4\cos^2(x) - 1 = -2\sin(x)\cos(x).$

Rearrange the terms:

$4\cos^2(x) + 2\sin(x)\cos(x) - 1 = 0.$

Now, we have a quadratic equation in $\cos(x)$. You can use the quadratic formula to solve for $\cos(x)$:

$\cos(x) = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.$

In this case, $A = 4$, $B = 2\sin(x)$, and $C = -1$. Plug these values into the quadratic formula and solve for $\cos(x)$. Once you have the values of $\cos(x)$, you can find the corresponding values of $\sin(x)$ using the Pythagorean identity $\sin(x) = \pm \sqrt{1 - \cos^2(x)}$.