Solve for approximate values of x

Question

6sin²x-5cosx-2=0

Answer 1

Ok, so you can't solve this problem by simply solving the equation for a trig function and then taking the inverse of a trig function.  We need start by turning this into something we can solve (like a quadratic polynomial).

 

First I want to translate this equation so its all in terms of one trig functions, cos x.

 

To do that remember the basic (pythagorean) trig identity: cos²x + sin²x = 1

 

I'll solve this for sin²x = 1-cos²x

 

Now I'll substitute this result into the original equation

 

6sin²x-5cosx-2=0

 

6(1-cos²x)-5cosx-2=0

 

Lets simplify a little before we turn this into a quadratic polynomial

 

6-6cos²x-5cosx-2=0

-6cos²x-5cos²x+4=0

6cos²x+5cosx-4=0

 

This still doesn't look like a polynomial.  To fix that we are going to define a function w=cosx

 

Substituting this back into the equation we get

 

6w²+5w-4=0

 

We can now solve this problem by factoring the polynomial.  I multiply the leading coefficient by the last coefficient to get 6*-4 = -24.  This factors into 8 and -3 which I can use to rewrite the polynomial as

 

6w²+8w-3w-4=0

2w(3w+4)-1(3w+4)=0

(3w+4)(2w-1)=0

 

Setting each factor equal to zero we get w = -4/3 and w = 1/2

 

Now we can resubstitute for w in terms of x to get

 

cosx=-4/3 and cosx=1/2

 

These are much easier to solve than the original problem.

 

Note that -1<=cosx<=1 which means that cosx=-4/3 will not yield any solutions.  So we need to focus on cosx=1/2

 

Taking the inverse cos of each side we get

 

cos^-1(cosx)=cos^-1(1/2) which means

 

x = cos^-1(1/2)

 

You can either plug this into your calculator or realize (from the unit circle) that cos(π/3)=1/2.

 

So x=π/3

 

But remember, based on the unit circle this isn't the onlly solution.  There is another solution at x=-π/3.  For ease, I want to represent both my solutions betwee 0 and 2π so the seocond solution is -π/3+2π=π5/3.

 

The last thing to remember is that the cosine is periodic with solutions at every 2π from the original solutions.  

 

So the complete solution is

 

x=π/3+2πN and 5/3π+2πN where N=0,+/-1,+/-2.+/-3.etc...

 

Good Luck.

 

Cheers

 

-Chris

Answer 2

If you replace sin² x by (1-cos²x), based on the Pythagorean Identity of Trigonometry, you have a quadratic equation in "cos x" which you can solve by various techniques learned in Algebra.

 

Then you'll have two(possible) answers for cos x, and then you can solve for x, in degrees, using arccos (whatever your cosine values are).  Remember to give values in all quadrants, not just first quadrants.