Trigonometry Question (just need to double check my answer..)

If 5cos[2(x+1)] = -7, then cos[2(x+1)] = -7/5 = -1.4

Given that cosine of EVERY angle is between -1 and 1 how did you find angle that solve cos[2(x+1)] = -1.4????!!!!
 
!!! I must've wrote the question wrong...my apologies...it should be:

7cos(2(x+1))=-5
 
Hyxie said:
This is how I solved it...would this be correct?
Click to expand...

I see. When you get your second solution, by subtracting the first from [MATH]2\pi[/MATH], this should be done to get a second [MATH]\theta[/MATH], not a second [MATH]x[/MATH].

[MATH]\theta_1 =2.3664[/MATH] (4 dp)

[MATH]\theta_2=2\pi-2.3664=3.9168[/MATH] (4 dp)

There are 2 different ways to finish up, but I'll assume you do the following:

Now, solve to find the two corresponding values of [MATH]x[/MATH]:
[MATH]2x+2=2.3664[/MATH][MATH]x=0.8132[/MATH] (4 dp)

[MATH]2x+2=3.9168[/MATH][MATH]x=0.9584[/MATH] (4 dp)

So [MATH]x_1=0.8132[/MATH], [MATH]x_2=0.9584[/MATH] are your two basic solutions. All others can be found by repeatedly adding and subtracting the period. However it is the period of [MATH]\text{cos}(2x+2)[/MATH] and that is [MATH]\pi[/MATH].
[MATH](\text{For cos}(ax+b) \text{, the period is }\frac{2\pi}{a}, \quad a≠0)[/MATH]
Now just keep adding and subtracting [MATH]\pi[/MATH] , keeping within the 'range': [MATH]-\pi≤x≤2\pi[/MATH]
-2.958, 0.183, 3.325
-2.183, 0.958, 4.1
(3 dp)
 
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