Question

1 The sum of the three smallest positive values of theta such that 4(cos*theta)(sin*theta) = 1 is k*pi. Find k?

Answer 1

`4costhetasintheta = 1`

`=>2costhetasintheta = 1/2`

`=>sin2theta = sin(pi/6)`

`=>2theta=npi+(-1)^n*(pi/6)`

`=>theta=1/2(npi+(-1)^n*(pi/6)) "where "ninZZ`

When `n=0totheta=pi/12`

When `n=1totheta=(5pi)/12`

When `n=2totheta=(13pi)/12`

By the condition of the problem

`pi/12+(5pi)/12+(13pi)/12=kpi`

So `k=19/12`

Answer 2

`k = (19)/12`

Explanation:

We can rewrite as

`2costhetasintheta = 1/2`

`sin(2theta) = 1/2`

Now I'm assuming that we're only consider values of `theta` greater than `0`.

`2theta = pi/6, (5pi)/6, (13pi)/6`

`theta = pi/12, (5pi)/12, (13pi)/12`

Now you just have to add the values up

`theta_"sum" = (19pi)/12`

Therefore `k = 19/12`

Hopefully this helps!

Answer 3

`k=19/12`

Explanation:

Here,

`4costhetasintheta=1`

`=>2sinthetacostheta=1/2`

`=>sin2theta==1/2 > 0=>I^(st) Quadrant orII^(nd)Quadrant`

`=>2theta=pi/6,(pi-pi/6),(2pi+pi/6),(3pi-pi/6),(4pi+pi/6),...`

`=>2theta=pi/6,(5pi)/6,(13pi)/6,(17pi)/6,(25pi)/6,(29pi)/6,...`

`=>theta=pi/12,(5pi)/12,(13pi)/12,(17pi)/12,...`

The sum of the three smallest positive values of theta `=kpi`

So,

`pi/12+(5pi)/12+(13pi)/12=kpi`

`=>(pi+5pi+13pi)/12=kpi`

`=>(19pi)/12=kpi`

`=>19/12=k`

`i.e. k=19/12~~1.5833`