Question

Sin2x+4(sinx+cosx)+4=0

Answer 1

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Answer 2

To solve the equation $\sin(2x) + 4(\sin(x) + \cos(x)) + 4 = 0$, we can follow these steps:

Step 1: Use trigonometric identities to simplify the equation. Step 2: Solve the resulting trigonometric equation.

Let's start with step 1:

1. Simplify the equation using trigonometric identities:

   We know that $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(x) = \sqrt{1 - \sin^2(x)}$.

   Substitute $\cos(x)$ in terms of $\sin(x)$: $\sin(2x) + 4(\sin(x) + \sqrt{1 - \sin^2(x)}) + 4 = 0$

Step 2: Solve the equation:

Let $u = \sin(x)$. Now the equation becomes: $2u + 4(u + \sqrt{1 - u^2}) + 4 = 0$

Simplify: $6u + 4\sqrt{1 - u^2} + 4 = 0$

Subtract 4 from both sides: $6u + 4\sqrt{1 - u^2} = -4$

Divide by 2: $3u + 2\sqrt{1 - u^2} = -2$

Move the $2\sqrt{1 - u^2}$ term to the other side: $3u = -2 - 2\sqrt{1 - u^2}$

Square both sides to eliminate the square root: $9u^2 = 4 + 8u - 4u^2$

Rearrange: $13u^2 - 8u - 4 = 0$

Now we have a quadratic equation in terms of $u$. We can solve this using the quadratic formula:

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

For $13u^2 - 8u - 4 = 0$, the coefficients are $a = 13$, $b = -8$, and $c = -4$.

Plug the values into the quadratic formula:

$u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 13 \cdot (-4)}}{2 \cdot 13}$

$u = \frac{8 \pm \sqrt{64 + 208}}{26}$

$u = \frac{8 \pm \sqrt{272}}{26}$

$u = \frac{8 \pm 16.49}{26}$

Now we have two possible values for $u$: $u_1 = \frac{24.49}{26}$ and $u_2 = -\frac{8.49}{26}$

However, since $u$ represents $\sin(x)$, which has a range of -1 to 1, the second solution is not valid.

So, $\sin(x) = \frac{24.49}{26}$.

Solve for $x$