Question

solve ( 2 cos ^ { 2 } ( t ) - 7 cos ( t ) + 5 = 0 ) for all solutions.
( t = ) where ( k in mathbb { z } )
question help: video

Explanation:

Step1: Let \(x = \cos(t)\)

The equation becomes \(2x^{2}-7x + 5=0\)

Step2: Factor the quadratic equation

We use the formula \(ax^{2}+bx + c=0\) (\(a = 2\), \(b=-7\), \(c = 5\)). The discriminant \(\Delta=b^{2}-4ac=(-7)^{2}-4\times2\times5=49 - 40=9\). Then \(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{7\pm3}{4}\). So \(x_1=\frac{7 + 3}{4}=\frac{10}{4}=\frac{5}{2}\), \(x_2=\frac{7-3}{4}=1\)

Step3: Re - substitute \(x=\cos(t)\)

Since \(- 1\leqslant\cos(t)\leqslant1\), \(\cos(t)=\frac{5}{2}\) is rejected. For \(\cos(t)=1\), we know that \(t = 2k\pi\), \(k\in\mathbb{Z}\)

Answer:

\(t = 2k\pi\)