Question

3 cos 2x - 5 sinx + 1 = 0.​

Answer 1

Ответ:$x+\left \{ {{30^{0}+k*360^{0} } \atop {150^{0} +k*360^{0} }} \right. ,k$∈Z

Объяснение:

Answer 2

To solve the equation $3\cos(2x) - 5\sin(x) + 1 = 0$, you can use trigonometric identities to simplify and solve for $x$.

First, we can use the double angle formula for cosine:

$\cos(2x) = 2\cos^2(x) - 1$

Now, replace $\cos(2x)$ in the equation:

$3(2\cos^2(x) - 1) - 5\sin(x) + 1 = 0$

Distribute and simplify:

$6\cos^2(x) - 3 - 5\sin(x) + 1 = 0$

Rearrange the terms:

$6\cos^2(x) - 5\sin(x) - 2 = 0$

Now, you can use the Pythagorean identity for sine:

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

So, we can rewrite the equation in terms of $\sin(x)$:

$6(1 - \sin^2(x)) - 5\sin(x) - 2 = 0$

Distribute and simplify:

$6 - 6\sin^2(x) - 5\sin(x) - 2 = 0$

Now, it's a quadratic equation in terms of $\sin(x)$. Let's make a substitution to simplify it further:

Let $u = \sin(x)$, then the equation becomes:

$6 - 6u^2 - 5u - 2 = 0$

Now, solve this quadratic equation for $u$:

$6u^2 + 5u - 8 = 0$

You can solve this quadratic equation using the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 6$, $b = 5$, and $c = -8$. Plug these values into the formula:

$u = \frac{-5 \pm \sqrt{5^2 - 4(6)(-8)}}{2(6)}$

Calculate the discriminant ($b^2 - 4ac$) first:

$5^2 - 4(6)(-8) = 25 + 192 = 217$

Now, continue solving for $u$:

$u = \frac{-5 \pm \sqrt{217}}{12}$

So, you have two possible solutions for $u$:

1. $u = \frac{-5 + \sqrt{217}}{12}$
2. $u = \frac{-5 - \sqrt{217}}{12}$

Now, remember that $u = \sin(x)$. To find $x$, you can use the arcsin (inverse sine) function:

1. x = \arcsin\left(\frac{-5 + \sqrt{217}}{12})
2. x = \arcsin\left(\frac{-5 - \sqrt{217}}{12})

These are the solutions for the equation $3\cos(2x) - 5\sin(x) + 1 = 0$. Depending on the range of $x$ you're interested in, you may need to consider the restrictions on the inverse sine function's range.