WMA12 Jan 2023

Question # 08

In this question you must show all stages of your working.

Solutions based entirely on calculator technology are not acceptable.

• Solve, for​​ π-2 < x < π, the equation​​ 

5sin (3x + 0.1) + 2 = 0

giving your answers, in radians, to 2 decimal places.​​ 

(4)​​ 

• Solve, for​​ 0 < θ < 360°, the equation​​ 

2tan θ sin θ = 5 + cos θ 

giving your answers, in degrees, to one decimal place.​​ 

(5)

SOLUTION

i- As the angle is​​ 3x + 0.1, the interval is

π-2 < x < π

3π-2 + 0.1≤3x + 0.1<3π + 0.1

-4.612<3x + 0.1<9.525

On solving, it gives ​​ 

5sin⁡3x+0.1+2=0 

sin⁡3x+0.1=-25 

3x+0.1=sin-1⁡-25

=-0.4115…

Other angles are at​​ 180-θ​​ in case of sine graph.

3x+0.1=-π+0.4115

=-2.7301

3x+0.1=π+0.4115

=3.5531

3x+0.1=2π-0.4115

=5.8716

Thus, the three angles are

3x+0.1=-2.7301, 3.5531, 5.8716

So the values of x for all four angles are​​ 

x=-0.4115-0.13,  -2.7301-0.13,      3.5531-0.13,      5.8716-0.13

x=-0.17, -0.94, 1.15, 1.92

ii-​​ 

2tan⁡θ sin⁡θ=5+cos⁡θ

Use trignometric identities to simplify to write the entire equation in terms of​​ sin⁡θ​​ and​​ cos⁡θ.​​ 

2sin⁡θcos⁡θsin⁡θ=5+cos⁡θ

2sin2⁡θ=5cos⁡θ+cos2⁡θ

Now, apply the trignometric identity # 1 from the above flashcard for converting the entire equation either in terms of ​​ cos⁡θ​​ or​​ sin⁡θ. Here, it is suitable to convert the entire equation​​ cos⁡θ​​ as that would give us the quadratic equation.​​ 

21-cos2⁡θ=5cos⁡θ+cos2⁡θ

2-2cos2⁡θ=5cos⁡θ+cos2⁡θ

0=3cos2⁡θ+5cos⁡θ-2

cos⁡θ=13  & -2

Remember, ​​ -1<cos⁡θ<1​​ so we will reject the​​ cos⁡θ​​ as -2 because it less than -1.​​ 

cos⁡θ=13     

θ=cos-1⁡13

θ=70.528………..

Don’t forget to find the other value of angle as the interval is​​ 0< θ< 360°. So to find it consider the quadrant other than first one where the​​ cos⁡θ​​ value is positive. And, that is in the 4th​​ quadrant. Hence, we will find the second angle as follow​​ 

θ=360-70.528=289,471…

Hecne the angles are

θ=70.5o    ,     289.5o