Solve each equation for all real values of x.
7.) 3 cos 2x - 5 cos x = 1
8.) 2 sin^2 x-5 sin x + 2 = 0
9.)3 sec^2 - 4 = 0
10.) tan x (tan x-1) = 0
7)
since cos 2x = 2cos^2 x - 1, we have
6cos^2 x - 5cos x - 4 = 0
(2cos x + 1)(3cos x - 4) = 0
cos x = -1/2 --> x = 2pi/3 or 5pi/3
cos x = 4/3 --> no solution
8)
(sin x - 2)(2sin x - 1) = 0
sin x = 2 --> no solution
sin x = 1/2 --> x = pi/6 or 5pi/6
9)
3sec^2 x = 4
sec x = ±2/√3
or, more familiarly,
cos x = ±√3/2
x = pi/6, 5pi/6, 7pi/6, 11pi/6
10)
tan X = 0 --> x = 0, pi
tan x = 1 --> x = pi/4, 5pi/4
7.) To solve the equation 3 cos 2x - 5 cos x = 1 for all real values of x, you can use the double-angle and half-angle trigonometric formulas.
Let's start by applying the double-angle formula for cosine: cos 2x = 2 cos^2 x - 1. Substituting this into the equation gives us:
3(2 cos^2 x - 1) - 5 cos x = 1
Expanding and rearranging the equation yields:
6 cos^2 x - 3 - 5 cos x = 1
6 cos^2 x - 5 cos x - 4 = 0
Now, we have a quadratic equation in terms of cos x. Let's solve it by factoring:
(2 cos x - 1)(3 cos x + 4) = 0
Setting each factor equal to zero gives us two individual equations:
2 cos x - 1 = 0 --> cos x = 1/2
3 cos x + 4 = 0 --> cos x = -4/3
To find the values of x, we need to use the inverse cosine (arccos) function:
cos x = 1/2 --> x = arccos(1/2) --> x = π/3 + 2πn or x = 5π/3 + 2πn
cos x = -4/3 (cosine of x cannot exceed 1 or be less than -1, so this isn't a valid solution)
Therefore, the solution to the equation 3 cos 2x - 5 cos x = 1 is:
x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.
8.) To solve the equation 2 sin^2 x - 5 sin x + 2 = 0 for all real values of x, we can use factoring or the quadratic formula.
First, let's try to factor the quadratic equation:
(2 sin x - 1)(sin x - 2) = 0
Setting each factor equal to zero gives us two individual equations:
2 sin x - 1 = 0 --> sin x = 1/2
sin x - 2 = 0 --> sin x = 2
However, sin x cannot exceed 1 or be less than -1, so the equation sin x = 2 doesn't have any real solutions. Let's focus on the equation sin x = 1/2.
To find the values of x, we can use the inverse sine (arcsine) function:
sin x = 1/2 --> x = arcsin(1/2) --> x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.
Therefore, the solution to the equation 2 sin^2 x - 5 sin x + 2 = 0 is:
x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.
9.) To solve the equation 3 sec^2 x - 4 = 0 for all real values of x, let's first simplify the equation.
First, we need to apply the reciprocal identity for secant: sec^2 x = 1/cos^2 x. Substituting this into the equation gives us:
3(1/cos^2 x) - 4 = 0
Rearranging the equation yields:
3/cos^2 x = 4
3 = 4cos^2 x
Dividing both sides by 4 gives us:
cos^2 x = 3/4
Taking the square root of both sides gives us:
cos x = ±√(3/4)
Since cosine can only be positive or negative, we have two possibilities:
cos x = √(3/4) --> x = ±arccos(√(3/4)) --> x = ±π/6 + 2πn or x = ±11π/6 + 2πn, where n is an integer.
cos x = -√(3/4) --> x = ±arccos(-√(3/4)) --> x = ±5π/6 + 2πn or x = ±7π/6 + 2πn, where n is an integer.
Therefore, the solution to the equation 3 sec^2 x - 4 = 0 is:
x = ±π/6 + 2πn, ±11π/6 + 2πn, ±5π/6 + 2πn, or ±7π/6 + 2πn, where n is an integer.
10.) To solve the equation tan x (tan x - 1) = 0 for all real values of x, we can set each factor equal to zero and solve separately.
tan x = 0 --> x = nπ, where n is an integer.
tan x - 1 = 0 --> tan x = 1 --> x = π/4 + πn or x = 5π/4 + πn, where n is an integer.
Therefore, the solution to the equation tan x (tan x - 1) = 0 is:
x = nπ, π/4 + πn, or 5π/4 + πn, where n is an integer.