Trigonometry! #61

(Sin^2 q - 1)^2 = 2

Sin q = +/- Sqrt [1 +/- Sqrt (2)]

By which no real angle between 0 and 2 Pi to suit for out of range for real numbers. Coming under [0, 2 Pi] means the number of solutions is 0.

Note: Trigonometric functions can manipulate complex number but there is no complex angle. Although they can operate complex number, it doesn't imply for a validity of complex angle! Just like a doctor can save many lives, that doesn't imply that every life being saved are equal or exactly the same thing.

Solving gives ±√(1±√2), all 4 solutions belong to set -1>x>1, while solution set of sintheta is -1<x<1, taking intersection of two sets, we get null set, therefore the equation has no solution .

If you solve it as a quadratic in $sin² \theta$ , you will get two solutions, one which is greater than 1 and the other being negative. Since both of these cannot be values that $sin ²\theta$ can take, hence no solution exists.

$\sin^{4}\theta-2\sin^{2}\theta-1=0$

$\sin^{2}\theta=\frac{2\pm\sqrt{4-4(1)(-1)}}{2(1)}$

$\sin^{2}\theta=\frac{2\pm2\sqrt{2}}{2}$

$\sin^{2}\theta=1\pm\sqrt{2}$

$\sin\theta=\sqrt{1\pm\sqrt{2}}$

As all four of these values lie outside the range $[-1,1]$ , this equation has no solutions.