Solve the equation 2cos^2x = 5sinx - 1 ...?

Question

Firstly, show that the equation may be written as 2sin^2x + 5sinx - 3 = 0, showing working

...

Then, solve the equation 2cos^2x = 5sinx - 1, in the range 0 to 360, showing working again

Thanks! x

Answer 1

Given 2cos^2x = 5sinx - 1, remember that cos^2(x) = 1 - sin^2(x) this gives:



2(1 - sin^2(x)) = 5sin(x) - 1 -----> 2 - 2sin^2(x) = 5sin(x) - 1 -----> 2sin^2(x) + 5sin(x) - 3 = 0



Now factor this equation. To help, let y = sin(x) and get:



2y^2 + 5y - 3 = 0



2y^2 - y + 6y - 3 = 0 -----> y(2y - 1) + 3(y - 1) = 0 -----> (2y - 1)(y + 3) = 0 -----> y = 1/2 or y = -3



Now go back to y = sin(x):



sin(x) = 1/2



or



sin(x) = -3



sin(x) has no real values when sin(x) = -3



When sin(x) = 1/2, then x = 30 or x = 150.



Now as an aside, if you let x = -π/2 + iln(3+√8), a complex number, then the equation works as well. You get this solution from sin(x) = -3. Typically though we restrict ourselves to the real numbers.

Answer 2

Substitute 2(1 - sin²(x)) for 2cos²(x)



2(1 - sin²(x)) = 5sin(x) - 1



Move everything to the left side:



2 - 2sin²(x)) - 5sin(x) + 1 = 0



Multiply -1



2sin²(x) + 5sin(x) - 3 = 0



let u = sin(x)



2u² + 5u - 3 = 0



Use the quadratic formula:



u = {-b ±sqrt[b² - 4(a)(c)]}]/2a



u = -5/4 ± sqrt[25 - 4(2)(-3)]/4



u = -5/4 ± 7/4



u = -12/4 and u = 2/4



u = -3 and u = 1/2



Reverse the substitution for u:



sin(x) = -3 and sin(x) = 1/2



The first root must be discard, because sin(x) can never be less than -1.



sin(x) = 1/2 at two well known points x = π/6 and x = 5π/6

Answer 3

Rewrite the cos^2 x as 1 - sin^2(x)



2(1 - sin^2(x)) - 5 sin(x) + 1 = 0

2 - 2sin^2(x) - 5sin(x) + 1 = 0

3 - 2sin^2(x) - 5sin(x) = 0



2sin^2(x) + 5sin(x) - 3 = 0



To solve, factor the expression (2 sin(x) - 1)(sin(x) + 3) = 0



Set each factor to 0 and solve: 2 sin(x) - 1 = 0 --> sin(x) = 1/2 x = 30, 150

sin(x) + 3 = 0 --> sin(x0 = -3 impossible

Answer 4

Theres a rule for this for addition log a + log b= log ab so u can rewrite as log5 (2x-a million)(2x+a million) = log5 (4x^2-a million)=a million so then you relatively inverse log the two facets with base 5 so 5^(log5(4x^2-a million))=5^a million so then 4x^2-a million=5 so then 4x^2-6=0 4x^2=6 so x^2=3/2 so x= +/- sqrt (3/2)