Find the fundamental solutions to the following equation (i.e., the solutions that are between 0 ° and 360 ° ). Give your answers in degrees. 20 sec 2 ( θ ) − 15 tan ( θ ) sec ( θ ) − 18 = 0

Answer 1

Hi Genevieve N.,

20sec²(θ) - 15tan(θ)sec(θ) - 18 = 0

20/cos²(θ) - 15sin(θ)/cos²(θ) - 18 = 0;

[20 - 15sin(θ)]/cos²(θ) = 18;

20 - 15sin(θ) = 18cos²(θ);

20 - 15sin(θ) = 18[1 - sin²(θ)];

20 - 15sin(θ) = 18 - 18sin²(θ);

18sin²(θ) - 15sin(θ) + 2 = 0;

[3sin(θ) - 2]*[6sin(θ) -1] = 0;

sin(θ) = 2/3 and sin(θ) = 1/6;

θ = sin^-1(2/3) = 41.8^°, and, θ = sin^-1(1/6) = 9.6^°

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And sin(θ) is positive in quadrant 1 and 2, therefore:

θ = 180° - 41.8° = 138.2^° and θ = 180° - 9.6^° = 170.4^°

So θ = 9.6^°, 41.8^°, 138.2^°, and 170.4^°, for 0° ≤ θ ≤ 360^°.

I hope this helps, Joe.

Answer 2

Hi Genevieve!

To find the solutions to the equation 20sec^2(θ)-15tan(θ)sec(θ)-18=0, normally we would to try to use identities so that we have only one trigonometric function in the equation (in this equation, say, any identity that would convert tangent to secant or vice versa). It is impossible here with the term -15tan(θ)sec(θ)--there is no tangent or secant identity that does anything with this term, so the next best try here would be to convert the existing trigonometric functions into sine and cosine and see what happens.

sec(θ)=1/cos(θ) and tan(θ)=sin(θ)/cos(θ), so if we plug these identities into the equation we get:

20(1/cos(θ))^2-15(sin(θ)/cos(θ))(1/cos(θ))-18=0

We can simplify the equation a bit by squaring 1/cos(θ) and doing the multiplication problems with fractions in the equation to get:

20/cos^2(θ)-15sin(θ)/cos^2(θ)-18=0

Notice that we have two fractions in the equation with the same denominator, cos^2(θ). Furthermore, cos^2(θ) is related to sin(θ) by the Pythagorean identity cos^2(θ)=1-sin^2(θ). So, I am going to multiply everything in the equation by cos^2(θ) to get rid of the fractions:

20-15sin(θ)-18cos^2(θ)=0

Then, after applying the Pythagorean identity cos^2(θ)=1-sin^2(θ), the equation becomes:

20-15sin(θ)-18(1-sin^2(θ))=0

The equation now has one trigonometric function, sin(θ). If I now substitute t=sin(θ), I will end up with a standard quadratic equation in t.

20-15t-18(1-t^2)=0

Distribute the -18 and simplify the equation:

20-15t-18+18t^2=0

18t^2-15t+2=0

It is simplest to use the quadratic formula t=(-b±√(b^2-4ac))/2a to solve this equation.

t=(-(-15)±√((-15)^2-4(18)(2))/(2(18))

t=(15±√(225-144))/36

t=(15±√(81))/36

t=(15±9)/36

If to the 15 we add and subtract 9 and then do the division by 36, we get the two solutions t=1/6 and t=2/3. Since t=sin(θ), we have sin(θ)=1/6 and sin(θ)=2/3. We will end up with four total solutions for θ. Notice that since we have sin(θ) equal to two positive numbers, the solutions will be in Quadrant I and Quadrant II where sine is positive. The Quadrant I solutions can be found by calculating arcsin(1/6) and arcsin(2/3), giving (to the nearest tenth) θ=9.6º and θ=41.8º. The Quadrant II solutions are mirror image angles of the ones we just found, and they are calculated by subtracting them from 180º. So θ=180º-9.6º=170.4º and θ=180º-41.8º=138.2º.

We can check these solutions by plugging them into the original equation to verify that they indeed work (all four give an answer very close to zero; since they are rounded this is what should happen). So the four solutions are 9.6º, 41.8º, 138.2º, and 170.4º.