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How can I solve the following trig equation: $$\sin x+2\cos x=2$$

I tried dividing by $\cos$ but it doesn't help much.

I also have the following question: $\sin x+p\cos x=2p$ how much should $p$ be so that this equation has solutions. I know it's related to the first one but I can't figure them out.

In the same chapter I have $\sin^2x+tg^2x=3/2$. I tried solving it with Weierstrass substitution but got to some really complicated equation is there an easier way?

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  • $\begingroup$ Well, one solution is $x=0+2n\pi$. $\endgroup$
    – lulu
    Jul 3, 2017 at 13:49
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    $\begingroup$ Please only post one question per post. If you believe they are related, you can link them back to each other but, as your question stands, there is no one answer which will suffice $\endgroup$
    – lioness99a
    Jul 3, 2017 at 13:51
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    $\begingroup$ There's a second one? In general, expressions of the form $a\sin x + b \cos x$ can always be rewritten as $Acos(x+c)$ for suitable amplitude $A$. If you do that in your case, the rest follows at once. $\endgroup$
    – lulu
    Jul 3, 2017 at 13:56
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    $\begingroup$ @Lola, Avoid squaring. Use $$\sin A=2(1-\cos A)$$ Now apply double angle formula $\endgroup$ Jul 3, 2017 at 15:51
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    $\begingroup$ OR use en.wikipedia.org/wiki/… $\endgroup$ Jul 3, 2017 at 15:52

6 Answers 6

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You can use the $\sin^2x+\cos^2x=1$ in the solution as follows:

$$ \sin x+2\cos x=2 \rightarrow 2\cos x=2-\sin x\\ \rightarrow (2\cos x)^2=(2-\sin x)^2 \\ \rightarrow 4 \cos^2 x=4+\sin^2 x -4 \sin x \\ \rightarrow 4(1-\sin^2 x)=4+\sin^2 x -4 \sin x \\ \rightarrow 4-4\sin^2 x = 4+\sin^2 x -4 \sin x \\ \rightarrow 5\sin^2 x- 4\sin x =0 \rightarrow \sin x (5\sin x -4)=0 \\ \rightarrow \sin x =0 \text{ or } \sin x =4/5 $$

From these $x$s can be found, e.g. $\sin x =0 \rightarrow x=\pm2\pi k, \pi\pm2\pi k$ and $\sin x =4/5 \rightarrow x=\arcsin (4/5)\pm2\pi k, \pi-\arcsin (4/5)\pm2\pi k$

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  • $\begingroup$ You can use \implies instead of \rightarrow as it is more accurate, mathematically. Though, it doesn't affect the accuracy of your answer, +1. $\endgroup$ Jul 3, 2017 at 14:06
  • $\begingroup$ @JaideepKhare Thanks for your comment. Henceforth I will use \implies. $\endgroup$
    – tempx
    Jul 4, 2017 at 6:23
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There is a standard method for solving linear equations in sine and cosine. Set $X=\cos x$ and $Y=\sin x$; now you get $$ \begin{cases} X^2+Y^2=1 \\[4px] Y+2X=2 \end{cases} $$ From the second equation you get $Y=2-2X$; substituting in the first equation you get the resolvent equation $$ 5X^2-8X+3=0 $$ so $$ \begin{cases} X=3/5\\ Y=4/5 \end{cases} \qquad\text{or}\qquad \begin{cases} X=1\\ Y=0 \end{cases} $$ Now you can determine $x$: $$ x=\arccos\frac{3}{5}+2k\pi=\arcsin\frac{4}{5}+2k\pi \qquad\text{or}\qquad x=2k\pi $$

The second question is indeed related: do the same substitution and find where the discriminant of the resolvent equation is $\ge0$.


About $\sin^2x+\tan^2x=3/2$, use the fact that $$ \tan^2x=\frac{1}{\cos^2x}-1 $$ so you can rewrite the equation as $$ 1-\cos^2x+\frac{1}{\cos^2x}-1=\frac{3}{2} $$ that leads to $$ 2\cos^4x+3\cos^2x-2=0 $$ This is a biquadratic and it yields $$ \cos^2x=\frac{1}{2} $$ (the negative solution should be discarded).

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  • $\begingroup$ Thank you for the answer,in my textbook the answer is written with arctan, any idea of how to write it like that? $\endgroup$
    – Lola
    Jul 3, 2017 at 14:34
  • $\begingroup$ @Lola If $\cos x=3/5$ and $\sin x=4/5$, then $\tan x=4/3$ and $0<x<\pi/2$, so you get $x=\arctan\frac{4}{3}$. But other equivalent expressions are possible. $\endgroup$
    – egreg
    Jul 3, 2017 at 14:43
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We need to solve $$\frac{1}{\sqrt5}\sin{x}+\frac{2}{\sqrt5}\cos{x}=\frac{2}{\sqrt5}$$ or $$\sin\left(x+\arcsin\frac{2}{\sqrt{5}}\right)=\frac{2}{\sqrt5},$$ which gives

$x=2\pi k$ or $x=\pi-2\arcsin\frac{2}{\sqrt{5}}+2\pi k,$ where $k\in\mathbb Z$.

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Here is another method which works most of the times when the equation contains $\sin x$ and $\cos x$ but no multiples or sub-multiples of $x$.

Let $\tan \frac{x}{2} = t$. (Observe that $x = (2k+1) \pi$ is definitely not a solution. So this substitution does not produce $\pm \infty$).

Then $\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$. Substitute this in your first equation to get a quadratic equation: $t(2t -1) = 0$. Thus $t = 0$ or $t = 1/2$. Thus, $x = 2n\pi$ or $x = 2 \arctan \frac{1}{2} + 2n\pi$.

For the second equation, use the same substitutions to get the quadratic equation: $3pt^2 - 2t + p = 0$. This equation will have real solutions iff the descriminant, $D = 4 - 12p^2 \geq 0 \iff -\frac{1}{\sqrt{3}} \leq p \leq \frac{1}{\sqrt{3}}$.

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To answer your first two questions, try using the following trick. Start with $\sin x + 2 \cos x = 2$. Get $\sin x$ by itself and square:

$$ 1 - \cos^2 x = (\sin x)^2 = 4(1-\cos x)^2 = 4 - 8\cos x + 4 \cos^2 x$$

or

$$ 5 \cos^2 x - 8 \cos x + 3 = 0 .$$

This is a quadratic in $\cos x$, which you can solve with the quadratic formula. Be sure to check that no extraneous roots were introduced.

The third question will be much harder as it involved trigonometric and linear terms.

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Another method is to express the trig functions in exponential form, then $\exp(ix)(2i+1)+\exp(-ix)(2i-1)=4i$. Multiplying by $\exp(-ix)$ produces a quadratic in $\exp(-ix)$ with solutions $\exp(-ix )= 1 $ and $(3-4i )/5$. The second solution gives $x = \tan^{-1}(4/3)$.

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