# √2 cos x - 1 = 0

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.

Trigonometric Equations

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