Trigonometry Question

Question

Solve 5sin²θ-18sinθ-8=0 for 0≤θ≤360°

a) θ≈203.6° or 336.4°

b) θ≈240° or 300°

c) θ≈33.6° or 156.4°

d) θ≈30° or 150°

Answer 1

I often find it easier to write the equation this way 5x^2 -18x-8=0 understanding that x is really sin x

(5x+2)(x-4)=0. X=4. This is extraneous as the range of sin x is from -1 to +1

5x=-2 x=-.4 gives you sinx =-.4 Take inverse sin to get x= -23.58 degrees which is the same as + 336.42 which is in quadrant 4

Sin is also negative in quadrant 3 so this would be 180+23.56= 203.58

Answer 2

The first step is to solve for sinθ by factoring the equation.

5sin²θ-18sinθ-8=0

(5sin²θ-20sinθ)+(2sinθ-8)=0

5sinθ(sinθ-4)+2(sinθ-4)=0

(5sinθ+2)(sinθ-4)=0

5sinθ+2=0 sinθ-4=0

sinθ=-2/5 sinθ=4

Then, recognize that sinθ=4 is an extraneous solution and use the inverse sine.

sin^-1(-2/5)=θ

-23.58=θ

Inverse sine will give negative answer when the angle is in the 3rd and 4th Quadrant.

So for the 3rd Quadrant answer, 180-(-23.58)=203.58

And the 4th Quadrant answer, 360+(-23.58)=336.42

So the answer is A.