Question

The number of principal solutions of tan⁡2θ = 1​

Answer 1

y=tan2θ

tan2θ=1

⟹2θ=tan

−1

(1)

⟹2θ=

4

π

+kπ, where k is a positive integer

Now,

for k=0⇒θ=

4

π

and for k=1⇒θ=

4

5π

0≤

4

π

≤2π and 0≤

4

5π

≤2π

∴2θ=

4

π

,

4

5π

⟹θ=

8

π

,

8

5π

Hence, the required principal solutions are

8

π