# Trigonometric equation with sine and cosine

So the equation is $3\cos ^2t + 5\sin t = 1$

Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$ which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$ Then I get $$-3t^2 + 5 t +2 = 0$$

Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^ 2 + y}$ where $y$ in this case will be $2/3$ ?

• Please clarigy your question. Why you used $t$ in both cases? – Michael Galuza Jul 26 '15 at 5:33
• your third and four steps are not good – Chiranjeev_Kumar Jul 26 '15 at 5:35
• I use t because it says so in my textbook. The question is solve the equation $3*\cos ^2t + 5*\sin t = 1$ and if the third and fourth step are bad then there should be another identity to use i guess.^ Thanks for your fast replies. – addde Jul 26 '15 at 5:36
• The original equation is expressed in terms of $t$, so if you are to make a substitution, it is a good idea to use a different variable, i.e. let $x=\sin(t)$, so you have got $-3x^2+5x+2=0$ – user265675 Jul 26 '15 at 5:40

$$3\cos ^2t + 5\sin t = 1$$

$$3(1-\sin ^2t) + 5\sin t -1 = 0$$

$$-3\sin ^2t + 5\sin t + 2 = 0$$

$$3\sin ^2t - 5\sin t - 2 = 0$$

Now , Let $\sin t =x$, then equation reduces to,

$$3x^2-5x-2=0$$

$$(3x+1)(x-2)=0$$, Now can you finish from here?

• Yes i can thanks for your help – addde Jul 26 '15 at 5:51
• @addde: you are welcome :) – Chiranjeev_Kumar Jul 26 '15 at 5:53

you have $$-3\sin(t)^2+5\sin(t)+2=0$$ let $$\sin(t)=u$$ then we have $$-3u^2+5u+2=0$$ divided by $-3$ gives $$u^2-\frac{5}{3}u-\frac{2}{3}=0$$ solving this equation we obtain $$u_{1,2}=\frac{5}{6}\pm\sqrt{\frac{25}{36}+\frac{24}{36}}$$ from here you will come to the result.

We have, $$3\cos ^2t + 5\sin t = 1$$ $$\implies 3(1-\sin ^2t) + 5\sin t = 1$$ $$\implies 3\sin ^2t+ 5\sin t-2=0$$ Factorizing the expression, we get $$(3\sin t-1)(\sin t+2)=0$$ $$\text{if}\ 3\sin t-1=0 \implies \sin t=\frac{1}{3}$$$$\implies \color{blue}{t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{Or} \ \color{blue}{t=(2n+1)\pi-\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{if}\ \sin t+2=0 \implies \sin t\neq -2$$ Hence, the general solution is $t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)$ or $t=(2n+1)\pi-\sin^{-1}\left(\frac{1}{3}\right)$

Where, n is any natural number