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

asked by guest
on Mar 25, 2025 at 8:55 am



You asked:

Solve the equation sin(x)2+2sin(x)3=0{\sin\left( x \right)}^{2} + 2 \sin\left( x \right) - 3 = 0 for the variable xx.

MathBot Answer:

The principal value in π2xπ2- \frac{\pi}{2} \le x \le \frac{\pi}{2} is x1.5707963x \approx 1.5707963


The real solutions are: {π(4n+1)2  |  nZ}\left\{\frac{\pi \left(4 n + 1\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\}


The complex solutions are: x={3π2ilog(3+22)+2πn  |  nZ}{6.2831853n+4.7123891.7627472i  |  nZ}x={3π2ilog(322)+2πn  |  nZ}{6.2831853n+4.712389+1.7627472i  |  nZ}\begin{aligned}x &= \left\{\frac{3 \pi}{2} - i \log{\left(3 + 2 \sqrt{2} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389 - 1.7627472 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{3 \pi}{2} - i \log{\left(3 - 2 \sqrt{2} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389 + 1.7627472 i\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}


ii is the imaginary unit, defined as i2=1i^2 = -1.