The Purplemath Forums

by **spycrab** on Mon Aug 23, 2010 12:36 am

Hi guys,
I have 2 equations on my homework that I am not allowed to use a calculator on, and need to find the exact answer/

$2sin^{2}(x)+5cos(x)=4$ where $0 \le x\le360$

And

$sin(2x)=2cos^{2}(x)$ where $0\le x\le 2PI$

**Sponsor**

by **stapel_eliz** on Mon Aug 23, 2010 11:25 am

spycrab wrote:I have 2 equations on my homework that I am not allowed to use a calculator on....

Why would you try to use a calculator for these? :confused:

Instead, try using the identities you've learned in trig and the factoring and solving techniques you learned back in algebra:

spycrab wrote: $2sin^{2}(x)+5cos(x)=4$

Apply a Pythagorean identity to convert the sine term to a cosine expression, solve the resulting quadratic-form equation for "cos(x)=", and then solve the two trig equations.

spycrab wrote: $sin(2x)=2cos^{2}(x)$

Convert the sine term using the double-angle formula, move one of the terms across the "equals" sign to join the other term, and factor the result. Then solve the two trig equations.

If you get stuck, please reply showing your work so far, starting with your conversions and factorizations.

Thank you! :wink:

The Purplemath Forums

solve 2sin^2(x) + 5cos(x) = 4, sin(2x) = 2cos^2(x)

solve 2sin^2(x) + 5cos(x) = 4, sin(2x) = 2cos^2(x)