Solve trigonometric equation by using the algebra technique, trig Pythagorean identities, inverse trig functions, and knowledge of the unit circle

Question

1. solve 2 sec x -4 = 0 for all radian values of x
2. find the solutions of cos x + 1 = - cos x in the interval [ 0, 2 pe]
3. solve 4 sin² x -3 = 0 for all degree values of x
4. solve sin x cos x = 3 cos x for all radian values of x

Answer 1

There are a lot of questions here. They're all done in a very similar manner to solving equations involving polynomials.

For example, to solve sin(x) + 1 = -sin(x) on [0,2π) you would take the following steps:

Add sin(x) to both sides to get 2sin(x) + 1 = 0

Subtract 1 from both sides to get 2sin(x) = -1

Divided both sides by 2 to get sin(x) = -1/2

Look on the unit circle to see sin(x) = -1/2 when x = 7π/6 or x = 11π/6

Answer 2

1. solve 2 sec x -4 = 0 for all radian values of x

Sec x = 2, Cos x = 1/2, x = π/3 or 60°

2. find the solutions of cos x + 1 = - cos x in the interval [ 0, 2 pe]

2 Cos x =-1, Cos x = -1/2, x = 2π/3 or 120°

3. solve 4 sin² x -3 = 0 for all degree values of x

Sin²x = 3/4, Sin x = (√3)/2, x = π/3 or 120°

4. solve sin x cos x = 3 cos x for all radian values of x

Cancel the cos x on both sides of the equation.

Sin x = 3 , there are no solutions because |Sin x| is always ≤1

Answer 3

1. answer is pi/3 or 5pi/3 + 2npi where n= any integer

2secx - 4 = 0

2secx = 4

secx =4/2 = 2

x = sec^-12 = 60 degrees = pi/3 radians + and - 2npi radians where n = any integer

or you could have rewritten secx as 1/cosx = 2, then cosx = 1/2 which is from the familiar 1-sqr3-2 right triangle, which shows up in part 1, 2, and 3 of this problem, but not part 4

the angle whose cosine is 1/2 is pi/3 radians or 5pi/3 radians

5pi/3 = 300 degrees.

x= pi/3 radians plus or minus any integer multiple of 2pi

and also 5pi/3 radians plus or minus any integer multiple of 2pi

x= pi/3 + 2npi, 5pi/3 + 2npi where n = any integer

2. Answer is 2pi/3 or 4pi/3 radians

cosx + 1 = - cosx

2cosx = -1

cosx = -1/2

x = cos^-1(-1/2) = 2pi/3 or 4pi/3 radians

= 120 or 240 degrees

those are the only solutions with 0<x<2pi

3. Answer is pi/3 or 4pi/3 + 2npi where n= any integer

4sin²x - 3 = 0

4sin²x = 3

sin²x = 3/4

sinx =+ or - (1/2) square root of 3

x = + or - pi/3 + or - any integer multiple of 2pi

and also 2pi/3 or 4pi/3 + or - any integer multiple of 2pi

x= pi/3 + 2npi, -pi/3 + 2npi, 2pi/3 + 2npi, and 4pi/3 + 2npi where n is any integer

4. Answer is x= pi/2 or 3pi/2 + 2npi radians where n= any integer

sinxcosx =3 cosx divide both sides by cosx which gives:

sinx = 3 which is not possible as sine is never greater than 1

there is no x such that sinx=3

But if cosx = 0, then x=pi/2 or 3pi/2 + 2npi where n = any integer

Then sinxcosx=3cosx will be (1)(0)=3(0) or 0=0

x=pi/2 or 3pi/2 + any integer multiple of 2pi will be solutions of sinxcosx=3cosx