Write the number of solutions of the equation tan x + sec x = 2 cos x

1.	Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π]
Answer» \(\frac{sinx}{cosx}+\frac{1}{cosx}\) = 2 x cos x sin x + 1 = 2 × (cos x)² sin x + 1 = 2 × (1 - (sin x)²) sin x + 1 = 2 – 2(sin x)² 2(sin x)² + sin x - 1 = 0 Consider a=sin x So, the equation will be 2a²+a -1 = 0 From the equation a=0.5 or -1 Which implies Sin x=0.5 or sin x=(-1) Therefore x=30° or 270° But for x=270° our equation will not be defined as cos (270° )=0 So, the solution for x = 30° According to trigonometric equations If sin x=sin a Then x = nπ – na Here sin x = sin30 So, x = nπ + (-1)ⁿ × 30 For n = 0, x = 30 and n =1,x = 150° and for n = 2,x = 390 Hence between 0 to 2π there are only 2 possible solutions.

Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π]

Answer»

\(\frac{sinx}{cosx}+\frac{1}{cosx}\) = 2 x cos x

sin x + 1 = 2 × (cos x)²

sin x + 1 = 2 × (1 - (sin x)²)

sin x + 1 = 2 – 2(sin x)²

2(sin x)² + sin x - 1 = 0

Consider a=sin x

So, the equation will be

2a²+a -1 = 0

From the equation a=0.5 or -1

Which implies

Sin x=0.5 or sin x=(-1)

Therefore x=30° or 270°

But for x=270° our equation will not be defined as cos (270° )=0

So, the solution for x = 30°

According to trigonometric equations

If sin x=sin a

Then x = nπ – na

Here sin x = sin30

So, x = nπ + (-1)ⁿ × 30

For n = 0, x = 30 and n =1,x = 150° and for n = 2,x = 390

Hence between 0 to 2π there are only 2 possible solutions.