Quadratic Trigonometric Equations
We have already mentioned that trigonometric ratios
are inter related and hence it is possible to restrict the equations with a single trigonometric ratio. However, in most cases the simplification turns into square of a trigonometric ratio. Thus, the equation becomes a quadratic equation in a single trigonometric equation. Hence, we need to follow the steps that are used in solution to quadratic equations
. That is either by the method of factorization or by use of quadratic formula. It will be very convenient to substitute a variable for the trigonometric ratio and solve the quadratic. After it is done, we can reverse the substitution and figure out the final answer. It is important that the solutions to the quadratic trigonometric equations are checked to score out any extraneous solution.
Let us consider the same equation sin x + cos x = 0 and try to solve in a different method.
sin x + cos x = 0
sin x = - cos x
squaring both sides, sin2 x = cos2 x
We know that,
cos2 x = 1 –
sin2 x and hence,
sin2 x = 1 – sin2 x
sin2 x = 1
sin2 x = 1/2
or, sin x = ±(√2)/2. Thus the solutions could be, x = (2n∏ + ∏/4), x = (2n∏ - ∏/4), x = (2n∏ + 3∏/4) and x = (2n∏ - 3∏/4). But the solutions x = (2n∏ + ∏/4) and x = (2n∏ - 3∏/4) do not satisfy the given equation.
Hence the actual solutions are only x = (2n∏ + 3∏/4) and x = (2n∏ - ∏/4) , where’ ‘n is any integer. This agrees with the solutions we figured out earlier.
In some cases, the interval of the actual angle may be given. In such cases, it is sufficient to find only the particular solutions within the given interval.
We will illustrate different types of quadratic trigonometric equations in the worked out examples.
Trigonometric Equations Examples Example 1: solve for all possible values of ‘x’ from the equation 1 + csc x = 3, ∏/2 < x < ∏
1 + csc x = 3
csc x = 3 – 1 = 2
Therefore, sin x = 1/2
We arrived at the value of sin x is 1/2 and it is positive. We know the sine ratio of angles that lie in first and second quadrants are positive. Also we know that 1/2 is the value of the sine ratio for the reference angle ∏. Since the direction says that the angle is in second quadrant, we need to use the reference angle for the second quadrant. Hence, the particular solution is,
x = ∏ – (∏/6) = 5∏/6
Example 2: solve for all possible values of ‘x’ from the equation sin x + √3cos x = 2
The problem can be solved by expressing either sin x or cos x in terms of the other and simplifying. But the method is cumbersome. Instead, we can use a simpler method by remembering the sine and cosine values for reference angle of ∏/3 or ∏/6. Let us show how the reference angle of ∏/3 is used in this problem.
sin x + √3cos x = 2. Dividing the equation throughout by2,
(1/2)*sin x + (√3/2)cos x = 1
Expressing (1/2) and (√3/2) as the values of cos ∏/3 and sin ∏/3 respectively,
cos ∏/3*sin x + sin ∏/3*cos x = 1
The left side is in the form of sum formula of sine. That is,
sin (x + ∏/3) = 1.
Now the right side can be expressed as sin (2n∏ + ∏/2), where ‘n’ is any integer.
Therefore, sin (x + ∏/3) = sin (2n∏ + ∏/2)
or, (x + ∏/3) = (2n∏ + ∏/2)
or, x = (2n∏ + ∏/2 + ∏/3)
or, x = (2n∏ + ∏/6)
The same answer can be obtained by expressing (1/2) and (√3/2) as the values of sin ∏/6 and cos ∏/6 respectively and using difference formula of cosine.
Example 3: solve for all possible values of ‘x’ from the equation cos x + sec x = 2
cos x + sec x = 2
or, cos x + (1/cos x) = 2
or, cos2 x + 1 = 2cos x
cos2 x – 2cos x + 1 = 0
The equation is a quadratic equation in ‘cos x’. It also appears to be an equation with identical roots.
(cos x – 1)2 = 0
or, cos x – 1 = 0
or, cos x = 1
The general solution for the above relation is,
x = 2n∏, where ‘n’ is any integer.
Example 4: solve for all possible values (in approximate degrees) of ‘x’ from the equation sec2 x + 1 = 3tanx, 0 < x < 2∏
sec2 x + 1 = 3tanx
(1 + tan2 x) + 1 = 3 tan x
tan2 x – 3tan x + 2 = 0
(tan x – 2)(tan x – 1) = 0
So, tan x = 2 and tan x = 1
The reference angles for tan x = 2 and for tan x = 1 are 63.4 deg (approximate) and 45 deg. Since the values obtained are positive, the angles of solution lie in first and third quadrants.
That is, the final solutions in degrees are, x = 45, x = 180 + 45 = 225, x ≈ 63.4 and x ≈ 180 + 63.4 ≈ 243.4
Example 5: solve for all possible values of ‘x’ from the equation cot x + csc x = 1
cot x + csc x = 1
or, cot x – 1 = csc x
Squaring both sides, or, (cot x – 1)2 = csc2 x
or, cot2 x – 2cot x + 1 = csc2 x
or, cot2 x + 1 – 2cot x = csc2 x
csc2 x – 2 cotx =
csc2 x (since cot^2 x + 1 = csc^2 x)
or, -2cot x = 0,
or, cot x = 0.
This leads to a general solution x = any odd multiple of ∏/2. But for the given equation is not valid for all odd multiples of ∏/2. For example, if x = 3∏/2, cot (3∏/2) + csc (3∏/2) = -1 and not 1.
Therefore, the general solution can only be x = (2n∏ + ∏/2), where ‘n’ is any integer.
Example 6: solve for all possible values of ‘x’ from the equation 2sin2 x + 5 cos x = -1
2sin2 x + 5 cos x = -1
or, 2 – 2cos2 x + 5 cos x = -1
cos2 x + 5 cos x + 3 = 0
cos2 x – 5cos x – 3 = 0
The equation is a quadratic equation in ‘cos x’. By factoring,
cos2 x – 6cos x + cos x – 3 = 0
or, 2cos x(cos x – 3) + 1(cos x – 3) = 0
or, (cos x – 3)(2cos x + 1) = 0
Hence, cos x = 3 and cos x = -1/2.
Obviously cos x = 3 is an extraneous solution and hence the only solution is cos x = -1/2.
The sign is negative which means, the angle is in second and third quadrants. The principal angle for which cosine ratio is 1/2, is ∏/3. Therefore, the general solution is,
x = 2n∏ + (∏ - ∏/3) = ( 2n∏ + 2∏/3) and x = 2n∏ + (∏ + ∏/3) = ( 2n∏ + 4∏/3), where ‘n’ is any integer.