Question

Sin^2x - 4sinx - 5 = 0 Please answer this. I want specific step-by-step instructions. THANKS!

Answer 1

Answer:

$x=\frac{3\pi }{2} +2\pi n$

Step-by-step explanation:

Interesting problem. This almost looks likes a 2nd degree trinomial, so let's actually make it into one. Allow sin(x) to equal 'y' to get:

$y^2-4y-5=0$

What multiplies to equal -5 but adds to equal -4? That would be -5 and positive 1. Therefore, it factors to be:

$(y-5)(y+1)=0$

$y=-1$

$y=5$

But recall our substitution that 'y' equals sin(x). Therefore:

$sin(x)=-1$

$sin(x)=5$

The range of sine is between -1 and +1. It is impossible (in the real plane) for sin(x) to equal 5. You would have to introduce the complex plane to get all the answers. The only real solution we have here is when:

$sin(x)=-1$

Sine of what equals -1? That would be 3pi/2 plus any phase shift of 2npi, where 'n' is an integer:

$x=\frac{3\pi }{2} +2\pi n$