We will discuss about the general solution of the equation square root of 2 cos x minus 1 equals 0 (i.e., √2 cos x  1 = 0) or cos x equals 1 by square root of 2 (i.e., cos x = \(\frac{1}{√2}\)).
How to find the general solution of the trigonometric equation cos x = \(\frac{1}{√2}\) or √2 cos x  1 = 0?
Solution:
We have,
√2 cos x  1 = 0
⇒ √2 cos x = 1
⇒ cos x = \(\frac{1}{√2}\)
⇒ cos x = cos \(\frac{π}{4}\) or, cos ( \(\frac{π}{4}\))
Let O be the centre of a unit circle. We know that in unit
circle, the length of the circumference is 2π.
If we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), and 2π.
Therefore, from the above unit circle it is clear that the final arm OP of the angle x lies either in the first or in the fourth quadrant.
If the final arm OP lies in the first quadrant then,
cos x = \(\frac{1}{√2}\)
⇒ cos x = cos \(\frac{π}{4}\)
⇒ cos x = cos (2nπ + \(\frac{π}{4}\)), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Therefore, x = cos (2nπ + \(\frac{π}{4}\)) …………….. (i)
Again, if the final arm OP of the unit circle lies in the fourth quadrant then,
cos x = \(\frac{1}{√2}\)
⇒ cos x = cos ( \(\frac{π}{4}\))
⇒ cos x = cos (2nπ  \(\frac{π}{4}\)), Where n ∈ I (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Therefore, x = cos (2nπ + \(\frac{π}{4}\)) …………….. (ii)
Therefore, the general solutions of equation cos x = \(\frac{1}{√2}\) are the infinite sets of value of x given in (i) and (ii).
Hence general solution of √2 cos x  1 = 0 is x = 2nπ ± \(\frac{π}{4}\), n ∈ I.
11 and 12 Grade Math
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