Is 20cos^2(x)-cos(x)-1=0 ?

The answer to whether 20cos^2(x)-cos(x)-1=0 is False

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prove 20cos^2(x)-cos(x)-1=0

prove

20 cos^{2} (x) - cos (x) - 1 = 0

False

Solution steps

20 cos^{2} (x) - cos (x) - 1 = 0

Show that the two sides are not equal

For

20 cos^{2} (x) - cos (x) - 1 = 0

plug in

x = 0

20 cos^{2} (0) - cos (0) - 1 = 18

20 cos^{2} (0) - cos (0) - 1

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 20 \cdot 1^{2} - 1 - 1

Simplify

= 18

The two sides are not equal

\Rightarrow False

tan (\frac{y}{2}) = csc (y) - cot (y)

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

sec (x) \cdot cot (x) - sin (x) = cos (x) \cdot cot (x)

\frac{cos ( θ )}{sec ( θ ) + tan ( θ )} = 1 - sin (θ)

\frac{sec ( θ )}{tan ( θ ) + cot ( θ )} = sin (θ)