Is prove-2sin^2(x)+cos(x)+1=0 ?

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

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prove-2sin^2(x)+cos(x)+1=0

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

- 2 sin^{2} (0) + cos (0) + 1 = 2

- 2 sin^{2} (0) + cos (0) + 1

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 + 1

Simplify

= 2

The two sides are not equal

\Rightarrow False

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

csc (2 x) + cot (2 x) = \frac{1 + cos ( 2 x )}{sin ( 2 x )}

1 - 2 sin^{2} (t) = 2 cos^{2} (t) - 1

\frac{1}{cos ^{2} ( θ )} = sec^{2} (θ)

- cos (2 t) sin (2 t) + sin (2 t) cos (2 t) + 0 = 0

Is prove-2sin^2(x)+cos(x)+1=0 ?
The answer to whether prove-2sin^2(x)+cos(x)+1=0 is False