Can someone help me on these

1. $cos2x=sin3x$

2. $\sqrt{3}tanx-secx-1=0$

3. $sec3\theta=sec\theta$

4. $sin4x-sin2x=cos3x$

thanks

first one 2x+3x = 90, or pi/2. and same angle in the third quarter with the 2n pi

the seond one use the identity

tan^2 x + 1 = sec ^2 x

Originally Posted by hacker804
Can someone help me on these

1. $cos2x=sin3x$

2. $\sqrt{3}tanx-secx-1=0$

3. $sec3\theta=sec\theta$

4. $sin4x-sin2x=cos3x$

thanks
1.
\displaystyle \begin{align*} \cos{2x} &= \sin{3x} \\ 1 - 2\sin^2{x} &= 3\sin{x} - 4\sin^3{x} \\ 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1 &= 0 \end{align*}

Write \displaystyle \begin{align*} X = \sin{x} \end{align*} to get the polynomial \displaystyle \begin{align*} 4X^3 - 2X^2 - 3X + 1 = 0 \end{align*}.

It's easy to see that \displaystyle \begin{align*} X = 1 \end{align*} satisfies this equation, so \displaystyle \begin{align*} X - 1 \end{align*} is a factor, and long division gives

\displaystyle \begin{align*} 4X^3 - 2X^2 - 3X + 1 &= 0 \\ (X - 1)\left(4X^2 + 2X - 1\right) &= 0 \\ 4(X - 1)\left(X^2 + \frac{1}{2}X - \frac{1}{4}\right) &= 0 \\ 4(X - 1)\left[X^2 + \frac{1}{2}X + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{4}\right] &= 0 \\ 4(X - 1)\left[\left(X + \frac{1}{4}\right)^2 - \frac{5}{16} \right] &= 0 \\ 4(X - 1)\left[ \left( X + \frac{1}{4} \right)^2 - \left( \frac{\sqrt{5}}{4} \right)^2 \right] &= 0 \\ 4(X - 1)\left( X + \frac{1}{4} - \frac{\sqrt{5}}{4} \right)\left( X + \frac{1}{4} + \frac{\sqrt{5}}{4} \right) &= 0 \\ X - 1 = 0 \textrm{ or } X + \frac{1}{4} - \frac{\sqrt{5}}{4} \textrm{ or } X + \frac{1}{4} + \frac{\sqrt{5}}{4} &= 0 \\ X = 1 \textrm{ or } X = \frac{-1 + \sqrt{5}}{4} \textrm{ or } X &= \frac{-1 - \sqrt{5}}{4} \end{align*}

So you need to solve \displaystyle \begin{align*} \sin{x} = 1 \end{align*} (easy), as well as \displaystyle \begin{align*} \sin{x} = \frac{1 + \sqrt{5}}{4} \end{align*} and \displaystyle \begin{align*} \sin{x} = \frac{1 - \sqrt{5}}{4} \end{align*} (not so easy). All will have exact value answers though. This will help you.

√3 tan⁡x-sec⁡x-1=0
√3 sin⁡x/cos⁡x -1/cos⁡x -1=0
√3 sin⁡x-1- cos⁡x=0
√3/2 sin⁡x-1/2 cos⁡x=1/2 {2 is √((√3)^2+(-1)^2 )}
cos⁡〖π/6 sin⁡x-〗 sin⁡〖π/6〗 cos⁡x=sin⁡〖π/6〗
sin⁡〖(π/6-x)=〗 sin⁡〖π/6〗
π/6-x=nπ+(-1)^n π/6
x=π/6-nπ-(-1)^n π/6
here n is a integer