Please help? Solve #5sin^2theta -18sintheta -8 =0# for #0°<=theta <=360°#

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Answer 1 · 2018-02-28T21:44:22

a

#5sin^2x-18sinx-8=0#

Notice that this is just a quadratic equation in #sinx#.

Let #u=sinx#

Then:

#5u^2-18u-8=0#

We now solve for #u#:

Factor:

#(5u+2)(u-4)=0=>u=-2/5 and u=4#

But #u=sinx#

#:.#

#sinx=-2/5# and #sinx=4#

#sinx=4# is an erroneous solution here. #sinx# is always in the interval:

#[-1,1]#

So we only have:

#sinx=-2/5#

Taking arcsine:

#x=arcsin(sinx)=arcsin(-2/5)=>x=-23.6color(white)(888)#1 .d.p.

Positive angle is:

#360-23.6=336.4#

This is in quadrant IV. We also have an angle in quadrant III

#180 + 23.6=203.6#

So a is the correct answer here.