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

3 Answers
May 16, 2018

#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#

May 16, 2018

#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!

May 16, 2018

# 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#