2sin2 3x + sin 3x - 1 = 0 untuk 0° ≤ x ≤ 360°. Misalkan sin 3x = p, Jadi himpunan penyelesaiannya adalah {10°, 50°, 90°, 130°, 170°, 210°, 250°, 290°, 330°} iNmI. Precalculus Examples Solve for ? sinx=cosx Step 1Divide each term in the equation by .Step 3Cancel the common factor of .Step the common 4Take the inverse tangent of both sides of the equation to extract from inside the 6The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from to find the solution in the fourth 7Step write as a fraction with a common denominator, multiply by .Step the numerators over the common 8Step period of the function can be calculated using .Step with in the formula for absolute value is the distance between a number and zero. The distance between and is .Step 9The period of the function is so values will repeat every radians in both directions., for any integer Step 10Consolidate the answers., for any integer $\begingroup$ I thought this one up, but I am not sure how to solve it. Here is my attempt $$\sin x-\sqrt{3}\ \cos x=1$$ $$\sin x-\sqrt{3}\ \cos x^2=1$$ $$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$ $$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$ $$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$ $$2\cos x\cos x-\sqrt{3}\sin x=0$$ $2\cos x=0\Rightarrow x\in \{\frac{\pi }22n-1n\in\Bbb Z\}$ But how do I solve $$\cos x-\sqrt{3}\sin x=0$$ asked Nov 10, 2018 at 115 $\endgroup$ 4 $\begingroup$Hint at the very beginning divide both sides by $2$ and use the formula for the sin of difference of 2 arguments answered Nov 10, 2018 at 117 MakinaMakina1,4441 gold badge7 silver badges17 bronze badges $\endgroup$ 1 $\begingroup$ Hint $$\cos x - \sqrt{3}\sin x = 0 \Leftrightarrow \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{3} \Leftrightarrow \tan x = \frac{\sqrt{3}}{3}$$ Note You can divide by $\cos x$, since if the case was $\cos x =0$, it would be $\sin x = \pm 1$ and thus the equation would yield $\pm \sqrt{3} \neq 0$, thus no problems in the final solution, as the $\cos$ zeros are no part of it. answered Nov 10, 2018 at 117 gold badges29 silver badges86 bronze badges $\endgroup$ 8 $\begingroup$ Multiply by the conjugate $\cosx - \sqrt{3} \sinx\cosx + \sqrt{3} \sinx = 0$. Then we have $\cos^2x-3\sin^2x=0$. This is the same thing as $1-4\sin^2x=0$ or $\sinx=\pm \frac{1}{2}$. NOTE OF CAUTION This gives you the answers to both the question and its conjugate. You'd have to plug in and check which ones are the answers you're looking for. answered Nov 10, 2018 at 124 JKreftJKreft2321 silver badge7 bronze badges $\endgroup$ $\begingroup$ You can turn the equation to a polynomial one, $$s-\sqrt3 c=1$$ is rewritten $$s^2=1-c^2=1+\sqrt3c^2,$$ which yields $$c=0\text{ or }c=-\frac{\sqrt3}2.$$ Plugging in the initial equation, $$c=0,s=1\text{ or }c=-\frac{\sqrt3}2,s=-\frac12.$$ Retrieving the angles is easy. answered Nov 10, 2018 at 1025 $\endgroup$ $\begingroup$ It's intersting, I believe, to consider also this other method for solving any linear equation in sine and cosine provided that the argument is the same for both functions. Recall that cosine and sine are abscissa and ordinate of points on the circumference of radius $1$ and center in the origin of the axes. Solving your first equation, therefore, is equivalent to finding the interection points between straight line $$r Y-\sqrt 3 X = 1 $$ and the circumference $$\gamma X^2+Y^2 = 1.$$ This brings you the system $$ \begin{cases} Y-\sqrt 3 X = 1\\ X^2+Y^2 = 1. \end{cases} $$ Replacing $Y = \sqrt 3 X + 1$ in the second equation gives you the quadratic equation $$2X^2 +\sqrt 3 X =0,$$ and, from here, to the solutions $$X_1 = 0, Y_1 = 1$$ and $$\leftX_2 = -\frac{\sqrt 3}{2}, Y_2 = -\frac{1}{2}\right,$$ with a straightforward trigonometric interpretation. I leave you as an exercise to apply the same approach to the equation you propose $$\cos x -\sqrt 3 \sin x = 0.$$ answered Feb 23, 2019 at 2007 dfnudfnu6,4051 gold badge8 silver badges26 bronze badges $\endgroup$ 1 You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged .