Solve sin(θ) tan(θ) − cos(θ) = 1 for 0 ≤ θ ≤ 360?

#sinthetatantheta-costheta=1#

We know : #tantheta= (sintheta)/(costheta)#

So,

#=>sintheta*(sintheta)/(costheta)-costheta=1#

#=>(sin^2theta)/(costheta)-costheta=1#

#=(sin^2theta- cos^2theta)/(costheta)=1#

#=>-(cos^2theta-sin^2theta)=costheta#

Using addition formulae:

#=>-cos2theta=costheta#

#=>cos2theta+costheta=0#

#=>2cos((3theta)/2)cos((theta)/2)=0#

#=>cos((3theta)/2)cos((theta)/2)=0#

I'm kind of stuck in the last step. Please could anybody help? Thanks!

Edit:

#:.cos(3/2theta)=0# or #cos(1/2theta)=0#

Start with #cos(3/2theta)=0#

#3/2theta=90^@#

#theta=60^@, 300^@#
The second result is from #360^@-theta_1#, where #theta_1 is our first answer.

Take #cos(1/2theta)=0#

#1/2theta=90^@#

#theta=180^@, 180^@# (this is a repeated root).

#:. theta=60^@, 180^@, 300^@#

There are two main trig identities we will use:

#tantheta=sintheta/costheta#

#sin^2theta+cos^2theta=1#

Start:

#sintheta tantheta - costheta=1#

Using #tantheta=sintheta/costheta#

#sintheta sintheta/costheta - costheta=1#

#sin^2theta/costheta-costheta=1#

#sin^2theta/costheta-cos^2theta/costheta=1#

#(sin^2theta-cos^2theta)/costheta=1#

#sin^2theta-cos^2theta=costheta#

#sin^2theta-costheta-cos^2theta=0#

Notice how we nearly have a quadratic in #costheta#. All we need to do is to get rid of that #sin^2theta# then solve like a normal quadratic.

From #sin^2theta+cos^2theta=1 =>sin^2theta=1-cos^2theta#

#sin^2theta-costheta-cos^2theta=0#

Using #sin^2theta=1-cos^2theta#

#(1-cos^2theta)-costheta-cos^2theta=0#

#1-costheta-2cos^2theta=0#

#2cos^2theta+costheta-1=0#

This is a quadratic in #costheta#. From here, we can factorise directly, or make a substitution to make it easier. I will use a substitution to see the quadratic more easily#

Let #u=costheta#

#2u^2+u-1=0#

#(2u-1)(u+1)=0#

#u=1/2# or #u=-1#

Remember #u=costheta#?

#costheta=1/2# or #costheta=-1#

Take #costheta=1/2#

We get our first answer from doing #cos^-1# on our calculator (or in this case, knowledge of special angles). For the second answer, we do #360^@-theta_1# where #theta_1# is the first answer we got. The reason for this can be seen in symmetries in the cosine graph.

#costheta=1/2#

#theta=60^@, 300^@#

#costheta=-1#

#theta=180^@, 180^@#
This is a repeated root - look how 180 is on the line of symmetry.

#:. theta=60^@, 180^@, 300^@#

And for reference, the graph of #y=sinx tanx -cosx-1# is:

Here is another way to crack the Problem :

# sinthetatantheta-costheta=1#.

#:. sinthetatantheta=1+costheta=2cos^2(theta/2)#.

#:. 2sin(theta/2)cos(theta/2)tantheta=2cos^2(theta/2)...(ast)#.

If, #cos(theta/2)=0#, then, since #0 le theta/2 le 180#,

#theta/2=90 rArr theta=180#.

Now, if #cos(theta/2)!=0#, then dividing #(ast)# by #2cos^2(theta/2)!=0#,

# tan(theta/2)tantheta=1, or, #

# tan(theta/2)*(2tan(theta/2))/(1-tan^2(theta/2)}=1, i.e., #

#2tan^2(theta/2)=1-tan^2(theta/2)#.

#:. 3tan^2(theta/2)=1 rArr tan(theta/2)=+-1/sqrt3#.

Selecting #theta/2" from "[0,180], theta/2=30, 150#.

# rArr theta=60, 300#.

Altogether, #theta in {60,180,300}#.

Enjoy Maths.!