Welcome to another video on solving trig equations the goal of this video is to
solve trig equations using a variety of strategies if you look at this first
equation it may be difficult to know how to approach it
we may try to add these two fractions together by obtaining a common
denominator but what I think might be helpful is multiplying this second
fraction by the conjugate of the denominator let's see what this does
so this first fraction will stay the same in the denominator we're going to
have 1 minus sine squared theta
and we'll leave the numerator in factored form
now one minus sine squared theta is equal to cosine squared theta so let's
go ahead and do that substitution now
and now this second fraction simplifies there's a common factor of cosine theta
so this simplifies out this will be to the first power and now notice that we
have two fractions with a common denominator so we can go ahead and add
these two fractions together our denominator will be cosine theta and the
numerator will have one plus one that would be two and notice our sine theta
is also simplify out we have sine theta plus a negative sine theta that would be
zero so we have two over cosine theta equals four now to solve this equation
for cosine theta we may want to perform cross products so for cosine theta
equals two dividing by four
all this comes down to where is cosine theta equal to 1/2 having a cosine
function value of 1/2 should remind you of a 30-60-90 right triangle where the
cosine of a 60-degree angle is equal to 1/2 that tells us our reference angle
will be 60 degrees next cosine theta is positive in both the first and the
fourth quadrant since cosine theta involves X so let's go ahead and sketch
a 60 degree reference angle in the first and the fourth quadrants these two
angles will have a cosine function value of 1/2 therefore theta could be 60
degrees or PI over 3 radians or theta could also be 300 degrees or 5 PI over 3
radians so this equation was fairly involved remember one of those
strategies we were using to verify identities was to multiply by the
conjugate let's take a look at another
here we have secant squared X minus two tan x equals four well we can replace
secant squared X with tan squared X plus one let's go ahead and do that we have
an equation that looks like it's going to be in quadratic form so let's go
ahead and set it equal to zero and rearrange the order of these terms if we
subtract 4 on both sides we'll have minus three on the left and luckily this
is factorable Tan squared X is a perfect square now we
can place the factors of negative three in the second positions put a minus
three here and a plus one here our outer product is negative three
Tan X the inner product is positive one tan X which does sum to the middle term
so if this product is equal to zero then either tan X must equal negative one or
tan X must equal three having a tangent function value of negative one should
remind us of the 45-45-90 right triangle or the tangent of 45 degrees is equal to
positive one so since we want a function value of negative one we have to
determine which quadrants the tangent function would be negative it would be
negative in the second quadrant so we'll sketch a 45 degree reference angle here
and it would also be negative in the fourth quadrant remember tangent is y
over X
so here X could equal 90 plus 45 or 135 degrees or 3 PI over 4 radians or 315
degrees which would be 7 PI over 4 okay for 10x equals 3 we're going to have to
use our calculators because this has not come out to a nice reference angle
remember if we solve this equation for x by taking the inverse tangent of both
sides we would have x equals inverse tangent 3 so let's go to the calculator
now make sure we're in Radian mode we are so second tangent brings up the
inverse looks like we have approximately 1.25 radians
going back to our coordinate plane 1.25 radians would be in the first quadrant
well tangent is also positive in the third quadrant so if we sketch a
reference angle of 1.25 radians we can determine another angle in this interval
that has a tangent function value of 3 so that would be PI plus 1.25 radians
which is approximately 4 point 3 9 so it looks like we have four different
solutions for this equation I think we have time for one more ok on this last
problem we have two different trig functions both with double angles so
what we'll try to do is use these double angle identities and perform some
substitutions to make this easier to solve luckily there's only one identity
for sine 2x so let's go ahead and replace sine 2x with 2 sine X cosine X
now for cosine 2x there's three different identities to choose from the
one that stands out for me is this third one because notice we have cosine two x
equals two cosine squared X minus one notice there's a plus one in the
original equation so those two will simplify out if I use this identity so
let's go ahead and give it a try cosine 2 X is equal to 2 cosine squared X minus
1 and then plus 1 from the original equation so these will go out and now
let's go ahead and try to set this equation equal to 0
now this is factorable there's a common factor of 2 cosine X let's try factoring
that out
that's going to leave us with sine X minus cosine X equals zero now normally
this is going to be a problem having two different trig functions but let's go
ahead and see what we can do with it so we'll set each of these factors equal to
zero dividing by two we have cosine x equals zero now over here what I'm going
to do is move the cosine X to the other side and now I'm going to divide both
sides by cosine X remember sine x over cosine X is tan X and of course cosine X
divided by cosine X is one so we need to find the angles on this interval where
cosine x equals zero and ten x equals one remember on the unit circle cosine
theta equals x so we want the points on the unit circle of where the
x-coordinate would be zero those occur on the y-axis so on this interval we
would have PI over two radians and also three PI over two next having ten X
equal to one to remind us of a 45-45-90 right triangle because tangent 45
degrees is equal to one so what that tells us is our reference
angle is 45 degrees remember tangent is positive in the first quadrant and also
in the third quadrant so we'll sketch a 45 degree reference angle here as well
so the first angle would be PI over 4 radians and the second angle would be
225 degrees or 5 PI over 4 radians so we had a couple of non routine equations
here hopefully the explanations made sense and you found them somewhat
helpful I hope you have a good day