Exercise 468Algebra › Equations and radicalsHigh schoolAdded 3 Nov 2026

A trigonometric equation

#trigonometric equation#solving#periodicity

Statement

Solve in $[0,2\pi]$ the equation
$2\cos^2 x-3\cos x+1=0.$

Hint

Set $t=\cos x\in[-1,1]$ .

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Substitution. $2t^2-3t+1=0$ , i.e. $(2t-1)(t-1)=0$ , so $t=1/2$ or $t=1$ .

Case $\cos x=1$ : $x=0$ (in $[0,2\pi]$ ) or $x=2\pi$ .

Case $\cos x=1/2$ : $x=\pi/3$ or $x=5\pi/3$ .

Conclusion. $x\in\{0,\pi/3,5\pi/3,2\pi\}$ .