Solving trigonometric equations

Hello, etheen!

I'll do a few of them . . .

Quote:

Solve over Real numbers:

$a)\;5\sin^2\!t+\cos^2\!t-2\:=\:0$

Replace $\cos^2\!t$ with $1 - \sin^2\!t$

$5\sin^2\!t + (1 - \sin^2\!t) - 2 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!t - 1 \:=\:0 \quad\Rightarrow\quad 4\sin^2\!t \:=\:1$

. . $\sin^2\!t \:=\:\tfrac{1}{4} \quad\Rightarrow\quad \sin t \:=\:\pm\tfrac{1}{2} \quad\Rightarrow\quad t \;=\;\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-3mm] \frac{5\pi}{6} + 2\pi n \end{Bmatrix}$

Quote:

$b)\;\sec^2\!t+\tan t\:=\:7$

Replace $\sec^2\!t$ with $\tan^2\!t + 1$

$(\tan^2\!t + 1) + \tan t \:=\:7 \quad\Rightarrow\quad \tan^2\!t + \tan t - 6 \:=\:0 \quad\Rightarrow\quad (\tan t - 2)(\tan t + 3) \:=\:0$

We have: . $\begin{array}{cccccccccc}<br /> \tan t - 2 \:=\:0 & \Rightarrow & \tan t \:=\:2 & \Rightarrow & t \:=\:\arctan(2) + \pi n \\<br /> \tan t + 3 \:=\:0 & \Rightarrow & \tan t \:=\:\text{-}3 & \Rightarrow & t \:=\:\arctan(\text{-}3) + \pi n \end{array}$

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$c)\;\sqrt{\sin t}\:=\:2\sin t-1$

Square both sides: . $\left(\sqrt{\sin t}\right)^2 \:=\:(2\sin t - 1)^2 \quad\Rightarrow\quad \sin t \:=\:4\sin^2\!t - 4\sin t + 1$

. . $4\sin^2\!t - 5\sin t + 1 \:=\:0 \quad\Rightarrow\quad (\sin t - 1)(4\sin t - 1) \:=\:0$

We have: . $\begin{array}{ccccccc}<br /> \sin t - 1 \:=\:0 & \Rightarrow & \sin t \:=\:1 & \Rightarrow & t \:=\:\frac{\pi}{2} + 2\pi n \\ \\[-3mm]<br /> 4\sin t - 1 \:=\:0 & \Rightarrow & \sin t \:=\:\frac{1}{4} & \Rightarrow& t \:=\:\arcsin\left(\frac{1}{4}\right) + 2\pi n \end{array}$

Quote:

$d)\;\tan t-\sin t\cos t\:=\:0$

We have: . $\frac{\sin t}{\cos t} - \frac{\sin t \cos^2\!t}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin t - \sin t\cos t}{\cos t} \:=\:0$

. . $\frac{\sin t(1 - \cos^2\!t)}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin t\cdot\sin^2\!t}{\cos t} \:=\:0 \quad\Rightarrow\quad \frac{\sin^3\!t}{\cos t} \:=\:0$

. . $\sin^3\!t \:=\:0 \quad\Rightarrow\quad \sin t \:=\:0 \quad\Rightarrow\quad t \:=\:\pi n$