Is 2sin^2(x)-cos(x)-2=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 2sin^2(x)-cos(x)-2=0

prove

2 sin^{2} (x) - cos (x) - 2 = 0

False

Solution steps

2 sin^{2} (x) - cos (x) - 2 = 0

Show that the two sides are not equal

For

2 sin^{2} (x) - cos (x) - 2 = 0

plug in

x = 0

2 sin^{2} (0) - cos (0) - 2 = - 3

2 sin^{2} (0) - cos (0) - 2

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

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

= 2 \cdot 0^{2} - 1 - 2

Simplify

= - 3

The two sides are not equal

\Rightarrow False

csc (t) - sin (t) = cot (t) \cdot cos (t)

\frac{tan ( θ ) + 6}{sec ( θ )} = 6 cos (θ) + sin (θ)

sin (x) + 5 = 4

\frac{1}{csc ( x )} = \frac{sin ( x )}{1}

sin^{2} (x) - 1 = 2 tan (x) - 1