5. Solving Trigonometric Equations

Rearranging the above equation, we get:

`cos theta=1/2`

We know the following :

Since

`cos (pi/3) = 1/2`

and `cos θ` is positive in the first and fourth quadrants, we have:

`theta=pi/3`

or

`theta=2pi-pi/3=(5pi)/3`

So `theta=pi/3` or `theta=(5pi)/3`

Using graphing software, we draw the curve of y = 2 cos²x − sin x − 1 in the region 0 ≤ θ < 2π. Wherever the curve cuts the x-axis will be the solution for our equation.

We see from the graph that the solutions are approximately:

x = 0.5
x = 2.6
x = 4.7

For more accurate solutions, we would just zoom in on the graph.

In this next graph, I have zoomed in to the second root (the one near x = 2.6). We can see that this root is x = 2.618 correct to 3 decimal places.

We could continue to zoom in as close as we like to get the accuracy that we require.

If the problem involved θ only, we would expect 2 solutions; one in the first quadrant and one in the second quadrant.

But here our problem involves `2θ`, so we have to double the domain (θ values) to account for all possible solutions.

We proceed as follows:

We solve

`sin 2θ = 0.8` for 0 ≤ 2θ < 4π.

The reference angle is

`α = arcsin 0.8 = 0.9273`

So the values for 2θ will be in quadrants I, II, V, VI.

2θ = 0.9273, or  π − 0.9273, or   2π + 0.9273, or 3π − 0.9273

That is

`2θ = 0.9273, 2.2143, 7.2105, 8.4975`

But we need values for θ, not 2θ, so we divide throughout by 2:

`θ = 0.4637, 1.1072, 3.6053, 4.2488`

Are our answers correct? As usual, we will check by graphing the original expression:

We can see from the graph that our 4 values are reasonable, since these are the only 4 values that satisfy `sin 2θ = 0.8`.

Note: We can always check our solutions with calculator, but it is easy to "miss out" on some of the required values if we only use calculator.

Solving for cos θ gives us:

`cos theta=+-1/4`

If `cos alpha=1/4`, then the reference angle is α = 1.3181.

So for `cos theta=1/4`, we have θ in the first and 4th quadrants. So

`θ = 1.3181 or 4.9651`

For `cos theta=-1/4`, we have θ in the 2nd and 3rd quadrants. So

`θ = 1.8235 or 4.4597`

So ` θ= 1.3181, 1.8235,` ` 4.4597` ` or 4.9651` radians.

We can see from the graph of `y=cos^2theta-1/16` that our answer is correct:

Factoring the LHS:

`6 sin^2θ − sin θ − 1 = 0`

`(2 sin θ − 1)(3 sin θ + 1) = 0`

So either

`2 sin θ − 1 = 0`

`sin θ = 1/2`

θ will be in 1st and 2nd quadrants.

`θ = 0.52360, 2.6180` (that is, `pi/6` or `(5pi)/6`)

OR

`3 sin θ + 1 = 0`

`sin θ = − 1/3`

θ will be in 3rd and 4th quadrants.

`θ = 3. 4814, 5. 9433`

Checking our solution on a graph:

So `θ = 0.52360, 2.6180, 3. 4814, 5. 9433`

Earlier we learned that `cos(x/2)=+-sqrt((1+cos x)/2)`, so we have:

`+-sqrt((1+cos x)/2)= 1 + cos x`

Squaring both sides gives:

`(1+cos x)/2=(1+cos x)^2`

`(1+cos x)/2=1+2\ cos x+cos^2x`

`1+cos x=2+4\ cos x+2\ cos^2x`

So we have:

`2\ cos^2 \ x + 3\ cos x + 1 = 0`

`(2\ co\s x + 1)(cos x + 1) = 0`

Solving, we get

`cos x = − 0.5` or `cos x = − 1`

Now `cos x=-1/2` gives `x=(2pi)/3,(4pi)/3`.

However, on checking in the original equation, we note that

`"LHS"=cos((4pi)/3xx1/2)` `=cos((2pi)/3)` `=-1/2`

but

`"RHS"=1+cos (4pi)/3=+1/2`

So the only solution for this part is `x=(2pi)/3.`

Also, `cos x=-1` gives `x = pi`.

So the solutions for the equation are `x=(2pi)/3or pi.`

A check of the graph of `y=cos x/2-1-cos x` confirms these results:

(2π/3 ≈ 2.0944 and π ≈ 3.14).

`cot 2theta= 1/(tan 2 theta)`, so we have:

`tan 2 theta-1/(tan 2 theta)=0`

`tan^2 2θ = 1`

`tan 2θ = ± 1`

 

Since `0 ≤ θ < 2π`, we need to consider values of `2theta` such that `0 ≤ 2θ < 4π`. Hence, solving the above equation, we have:

`2 theta=pi/4,(3pi)/4,(5pi)/4(7pi)/4,` `(9pi)/4,(11pi)/4,` `(13pi)/4,(15pi)/4`

Dividing throughout by 2 gives us the full set of solutions in the required domain, `0 <= theta <2pi`:

`theta=pi/8,(3pi)/8,` `(5pi)/8,(7pi)/8,` `(9pi)/8,(11pi)/8,` `(13pi)/8,(15pi)/8`

`4\ tan x− sec^2x= 0`

Writing this in terms of `sin x` and `cos x` only:

`4(sin x)/(cos x)-1/(cos^2x)=0`

Multiplying throughout by `cos x`:

`4\ sin x\ cos x=1`

Dividing both sides by 2:

`2\ sin x\ cos x=1/2`

Recognizing the LHS is `sin 2x`, from before:

`sin 2x,=0.5`

In 0 ≤ x < 2π, we need to find values of 2x such that 0 ≤ 2x < 4π. (Twice the original domain.)

So the values for `2x` are:

`2x=pi/6,(5pi)/6,(13pi)/6,(17pi)/6`

Dividing throughout by 2 gives our required values for `x`:

`x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

or, in decimal form:

`x = 0.2618, 1.309, 3.403, 4.451`

We recognise the left hand side to be in the form:

`sin(a − b) =` ` sin a cos b − cos a sin b,`

where `a = 2x` and `b =x`.

So

`sin 2x\ cos x − cos 2x\ sin x`

`= sin(2x − x)`

`= sin x`

Now, we know the solutions of `sin x = 0` to be:

`x = 0, π`.

[Why?]

sin 4x − cos 2x = 0

2sin 2x cos 2x − cos 2x = 0

Factoring gives:

cos 2x (2 sin 2x − 1) = 0

EITHER

`cos 2x= 0`

`2x=pi/2,(3pi)/2,(5pi)/2,(7pi)/2`

`x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4`

OR

`sin 2x= 1/2`

`2x=pi/6,(5pi)/6,(13pi)/6,(17pi)/6`

`x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

Or, in decimal form:

`x= 0.26, 0.79, ` `1.31, 2.36, ` `3.40, 3.93, ` `4.45, 5.50.`

The graph of `y = sin 4x− cos 2x` is as follows. We can see from where the graph cuts the x-axis that our answers are reasonable.