solving trigonometry

Question

Solve $16 \cos^2 x + 6 \sin x = 17$ for $0 < x < 2\pi$

Steps:

$16(1-\sin^2x)+6\sin x = 17$

$16\sin^2x-6\sin x + 1 = 0$

let $y = \sin x$

$16y^2 - 6y + 1 = 0$

I was not able to solve this quadratic equation. It has no real roots. What was done incorrectly?

Your title says nothing about what your question really is. I think the first line of your question should have been the title too. — Steven Alexis Gregory, Sep 06 '21 at 14:29
Wolfram Alpha says the maximum value of $16\cos^{2}x+6\sin x$ for real $x$ is $256/16 < 17.$ See it here. If you were meant to also try complex $x$ then the condition $0 < x < 2\pi$ is ambiguous. Unless you copied the problem incorrectly, I don't think this is your fault. The error is in the problem itself. — David K, Sep 06 '21 at 23:00
@DavidK Actually, without the typo it is 265/16 or 16 + 9/16, easily obtained 9/16 from the OP's last quadratic's determinant, and adding 16 to get the original quadratic. — Dávid Laczkó, Sep 07 '21 at 20:47
That's right, WA actually said $265/16$ (still less than $17$), but I accidentally transposed two digits. — David K, Sep 07 '21 at 23:54

score 1 · Answer 1 · answered Sep 25 '21 at 13:20

As noticed your derivation is fine, the quadratic equation $16y^2 - 6y + 1 = 0$ has not real roots indeed $f(y)=16y^2 - 6y + 1$ represents a parabola that opens upward with vertex at $\left(\frac 3{16},\frac 7{16}\right)$.

As an alternative, we have that

$$16 \sin^2 x - 6 \sin x + 1=0 \iff 16 \sin^2 x +1= 6 \sin x $$

which can't have solutions for $\sin x \le 0$ and for $\sin x >0$ by AM-GM

$$16 \sin^2 x +1\ge 8 \sin x>6 \sin x$$

solving trigonometry

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