Question

If sinθ=k,0<k<1,θ∈Q1sinθ=k,0<k<1,θ∈Q1 , then tanθ=​

Answer 1

Answer:

We have, θ∈[0,37π]

=[0,2π+3π]

=[0,2π]+[2π,37π]

=T+[2π,37π] Where T stands for time period of the sinx.

Now sinθ achieves a minimum value of -1 once in its time period, and maximum value of 1, only once in its time period.

Now the given interval is 

T+[2π,37π]

In the interval of [2π,37π], sinx neither achieves 1 nor achieves -1.

Hence a=1 or a=−1 for the equation sinθ=a to have only one solution in the given interval of θ.