<p>Show that&nbsp;\(3\sin^2{x}-2\cos{x}=1\) can be written in the form \(a\cos^2{x}+b\cos{x}+c=0\) where&nbsp;\(a,b\) and \(c\) are integers.</p>
<p>Hence, solve \(3\sin^2{x}-2\cos{x}=1\) for&nbsp;\(0 \leq x \leq 360^{\circ}\).</p>

Show that \(3\sin^2{x}-2\cos{x}=1\) can be written in the form \(a\cos^2{x}+b\cos{x}+c=0\) where \(a,b\) and \(c\) are integers.

Hence, solve \(3\sin^2{x}-2\cos{x}=1\) for \(0 \leq x \leq 360^{\circ}\).

Trigonometric Equations

Trigonometry

International Baccalaureate Mathematics: Applications and Interpretation Standard Level

Trigonometry - Trigonometric Equations

Trigonometry - Trigonometric Equations