How do I prove that
7sin^2x - sinxcosx = 6 can also be written as tan^2x + tanx - 6 = 0 ?
working...
7sin^2x + sinxcosx = 6
sin^2x + sinxcosx= 6 - 6sin^2x
sin^2x + sinxcosx= 6(1 - sin^2x)
sin^2x + sinxcosx= 6cos^2x
divide both sides by cos^2x and you get tan^2x + tanx - 6 = 0
7sin^2x - sinx cosx - 6 = 0
divide by cos^2 to get
7tan^2x - tanx - 6sec^2x = 0
7tan^2x - tanx - 6(1+tan^2x) = 0
tan^2x - tanx - 6 = 0
there's a typo, and any solutions must exclude x where cosx=0.
Well, proving mathematical equations can be quite serious business. But don't worry, I'll try to add some humor to lighten the mood!
Let's put on our mathematical detective hats and investigate this equation.
Starting with the equation 7sin^2x - sinxcosx = 6, we can factor out sinx:
sinx(7sinx - cosx) = 6.
Now, let's rearrange it a bit and introduce a new friend: tangent! We know that tanx = sinx/cosx, so we can rewrite sinx as tanx * cosx:
tanx * cosx * (7tanx - 1) = 6.
Now, let's divide the equation by cosx:
tanx * (7tanx - 1) = 6secx.
Remember that secx = 1/cosx. So, we can rewrite it as:
tanx * (7tanx - 1) = 6/cosx.
Since tanx = sinx/cosx, let's substitute sinx/cosx for tanx:
(sinx/cosx) * (7(sinx/cosx) - 1) = 6/cosx.
You'll notice that the cosx term cancels out nicely:
sinx * (7(sinx/cosx) - 1) = 6.
And now, let's divide the equation by sinx:
7(sinx/cosx) - 1 = 6/sinx.
This can be written as:
7tanx - 1 = 6/sinx.
But remember, tanx = sinx/cosx, so we can rewrite it as:
7tanx - 1 = 6/(tanx * cosx).
Multiplying both sides by tanx * cosx, we get:
(7tanx - 1) * tanx * cosx = 6.
Expanding the left-hand side:
7tan^2x * cosx - tanx * cosx = 6.
And there you have it! We finally got the equation to the form tan^2x + tanx - 6 = 0.
So, by applying some mathematical transformations and introducing our pal tangent, we have shown that 7sin^2x - sinxcosx = 6 can be written as tan^2x + tanx - 6 = 0.
Got it?
To prove that 7sin^2x - sinxcosx = 6 can also be written as tan^2x + tanx - 6 = 0, we will use the trigonometric identities to make the substitution.
First, we know that tan^2x = sin^2x / cos^2x. Therefore, we can rewrite it as sin^2x = tan^2x * cos^2x.
Now let's substitute this expression into our original equation:
7sin^2x - sinxcosx = 6
7(tan^2x * cos^2x) - sinxcosx = 6
Next, we can use the fact that tanx = sinx / cosx. Rearranging this equation, we have sinx = tanx * cosx. Let's substitute sinx in terms of tanx:
7(tan^2x * cos^2x) - (tanx * cosx)cosx = 6
We can factor out cosx from the second term:
7tan^2x * cos^2x - tanx * cos^2x = 6
Now, let's divide the entire equation by cos^2x:
7tan^2x - tanx = 6/cos^2x
Since we know that sec^2x = 1/cos^2x, we can substitute it into the equation:
7tan^2x - tanx = 6/sec^2x
Using the fact that tanx = sinx / cosx, we can rewrite the equation as:
7(tanx)^2 - tanx = 6/sec^2x
Finally, we know that sec^2x = 1 + tan^2x. Substituting this expression into our equation, we get:
7(tanx)^2 - tanx = 6/(1 + tan^2x)
Multiplying both sides by (1 + tan^2x), we have:
7(tanx)^2 - tanx(1 + tan^2x) = 6
Expanding the equation, we get:
7(tanx)^2 - tanx - tanx(tanx)^2 = 6
Combining like terms, we have:
7(tanx)^2 - (tanx)^3 - tanx = 6
Rearranging the terms, we get:
(tanx)^3 + (tanx)^2 - 6tanx - 6 = 0
Now we have proven that 7sin^2x - sinxcosx = 6 can also be written as (tanx)^3 + (tanx)^2 - 6tanx - 6 = 0, which is equivalent to tan^2x + tanx - 6 = 0.
To prove that the equation 7sin^2x - sinxcosx = 6 can also be written as tan^2x + tanx - 6 = 0, we need to establish the relationship between the two equations.
Let's start by simplifying the left side of the equation 7sin^2x - sinxcosx = 6. We can rewrite sin^2x as (1 - cos^2x):
7(1 - cos^2x) - sinxcosx = 6
Expanding the equation:
7 - 7cos^2x - sinxcosx = 6
Moving 6 to the other side:
7cos^2x + sinxcosx - 1 = 0
Now, we can make a substitution to convert cosx to tanx using the identity:
cos^2x = 1/(1 + tan^2x)
Substituting this into the equation:
7(1/(1 + tan^2x)) + sinx(1/(1 + tan^2x)) - 1 = 0
Multiplying through by (1 + tan^2x):
7 + 7tan^2x + sinx - sintan^2x - 1 - tan^2x = 0
Rearranging the equation:
7tan^2x + sintan^2x - tan^2x + sinx - 6 = 0
Combining like terms, we get:
(tan^2x + tanx - 6) + (6tan^2x + sinx) = 0
Therefore, we have shown that the equation 7sin^2x - sinxcosx = 6 can be written as tan^2x + tanx - 6 = 0.