Question

solve ( 4 cos ^ { 2 } ( x ) - 19 cos ( x ) + 12 = 0 ) for all solutions ( 0 leq x < 2 pi ).

( x = )

give your answers as values accurate to at least two decimal places in a list separated by commas.

question help: video

Explanation:

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

The equation \( 4\cos^{2}(x)-19\cos(x)+12 = 0 \) becomes \( 4t^{2}-19t + 12=0 \).
Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for \( at^{2}+bt + c = 0 \), here \( a = 4 \), \( b=-19 \), \( c = 12 \).

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Step2: Solve for \( t \)

Case 1: \( t=\frac{19 + 13}{8}=\frac{32}{8}=4 \)
Case 2: \( t=\frac{19-13}{8}=\frac{6}{8}=0.75 \)
Since \( - 1\leqslant\cos(x)\leqslant1 \), \( \cos(x)=4 \) is rejected.

Step3: Solve \( \cos(x)=0.75 \)

\( x=\cos^{-1}(0.75)\approx0.72 \) or \( x = 2\pi-\cos^{-1}(0.75)\approx2\pi - 0.72=5.56 \)

Answer:

\( 0.72,5.56 \)