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# Need help on C4 Trigonometry Watch

Solve the equation 5cos2x - 3sinx = 4 for 0°≤ x ≤ 360°

I've put the equation into the form of Rcos(x+a) already. The answer I got was correct according to the text book and I've got this.

√34 cos (x + 30.96) = 4

I've then rearranged the formula into x + 30.96 = cos-1(4/√34).

This is the part I am stuck on for questions like this. How do I find the value of x? Apparently there are two answers, but I don't know what to do to find them.

2. (Original post by Jacklicy)

Solve the equation 5cos2x - 3sinx = 4 for 0°≤ x ≤ 360°

I've put the equation into the form of Rcos(x+a) already. The answer I got was correct according to the text book and I've got this.

√34 cos (x + 30.96) = 4

I've then rearranged the formula into x + 30.96 = cos-1(4/√34).

This is the part I am stuck on for questions like this. How do I find the value of x? Apparently there are two answers, but I don't know what to do to find them.

So x is just:

x = cos^-1(4/√34) -30.96

So work out cos inverse of (4/√34), which would give u the two values
3. (Original post by 2710)
So x is just:

x = cos^-1(4/√34) -30.96

So work out cos inverse of (4/√34), which would give u the two values
I've got 15.73 for one of the answers which is correct. How do I find the other one?
4. (Original post by Jacklicy)
I've got 15.73 for one of the answers which is correct. How do I find the other one?
So you managed to pass C1-3 without knowing how to find trig solutions?

Edit: see my later post.

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5. (Original post by Jacklicy)

Solve the equation 5cos2x - 3sinx = 4 for 0°≤ x ≤ 360°

I've put the equation into the form of Rcos(x+a) already. The answer I got was correct according to the text book and I've got this.

√34 cos (x + 30.96) = 4

I've then rearranged the formula into x + 30.96 = cos-1(4/√34).

This is the part I am stuck on for questions like this. How do I find the value of x? Apparently there are two answers, but I don't know what to do to find them.

5cos2x - 3sinx = 4

Use cos2x=1-2(sinx)^2

5(1-2(sinx)^2) -3sinx=4

5-10(sinx)^2 -3sinx = 4

-10(sinx)^2 -3sinx +1 = 0

Therefore sinx=-0.5,0.2

therefore x=330,210,11.54,168.46
6. (Original post by Jacklicy)

Solve the equation 5cos2x - 3sinx = 4 for 0°≤ x ≤ 360°

I've put the equation into the form of Rcos(x+a) already. The answer I got was correct according to the text book and I've got this.

√34 cos (x + 30.96) = 4

I've then rearranged the formula into x + 30.96 = cos-1(4/√34).

This is the part I am stuck on for questions like this. How do I find the value of x? Apparently there are two answers, but I don't know what to do to find them.

On further inspection, 5cos2x - 3sinx = 4 cannot be written in Rcos(x + a) form. Take a look at the x's.

Posted from TSR Mobile
7. (Original post by Jacklicy)
I've got 15.73 for one of the answers which is correct. How do I find the other one?
Look at the cos graph. You have cos^-1(4/√34) = roughly 47 degrees.

So which other degree also gives u cos 47? Looking at the graph, you will have 360 - 47

Do it with the precise figures
8. (Original post by Joshmeid)
On further inspection, 5cos2x - 3sinx = 4 cannot be written in Rcos(x + a) form. Take a look at the x's.

Posted from TSR Mobile
According to my text book it is correct.
9. (Original post by 2710)
Look at the cos graph. You have cos^-1(4/√34) = roughly 47 degrees.

So which other degree also gives u cos 47? Looking at the graph, you will have 360 - 47

Do it with the precise figures
Using the correct figures I ended with with 313.3. However my MEI text book reveals the second answer to be 282.4, which is pretty confusing.
10. (Original post by Jacklicy)
Using the correct figures I ended with with 313.3. However my MEI text book reveals the second answer to be 282.4, which is pretty confusing.
Thats correct, but then u need to minus the 30.96, which gives you your answer

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