Lesson Solving more complicated problems on trigonometric equations
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<H2>Solving more complicated problems on trigonometric equations</H2> In this lesson you will find the solutions of these trigonometric equations: 1. {{{2cos^2(x) + cos(x)}}} = {{{0}}}. 2. {{{2sin^2(x) -7sin(x) + 3}}} = {{{0}}}. 3. {{{2cos^2(theta)}}} = {{{sin(theta)+1}}}. 4. {{{3cos(beta) + 3}}} = {{{2sin^2(beta)}}}. 5. {{{cos(2x) + 6sin^2(x)}}} = {{{4}}}. <H3>Problem 1</H3>Solve an equation {{{2cos^2(x) + cos(x)}}} = {{{0}}}. <B>Solution</B> <pre> {{{2cos^2(x) + cos(x)}}} = {{{0}}}. (1) Factor the left side. You will get cos(x)*(2cos(x) + 1) = 0. The last equation splits in two separate equations: 1. cos(x) = 0 ---> x = {{{pi/2}}} and/or x = {{{-pi/2}}}. In the interval [{{{0}}},{{{2pi}}}) there are no other solutions. If you want to write general solution then x = {{{k*pi}}}, where k is any integer, k = 0, +/-1, +/-2 . . . 2. 2cos(x) + 1 = 0 ---> cos(x) = {{{-1/2}}} ---> x = {{{2pi/3}}} and/or x = {{{4pi/3}}}. In the interval [{{{0}}},{{{2pi}}}) there are no other solutions. If you want to write a general solution, then x = +/- {{{pi/3}}} + {{{(2k+1)*pi}}}, where k is any integer, k = 0, +/-1, +/-2 . . . <B>Answer</B>. The solution of the original equation (1) is the union of the solutions to n.1 and n.2. </pre> <H3>Problem 2</H3>Solve an equation {{{2sin^2(x) - 7sin x + 3}}} = {{{0}}} for all real values of x. <B>Solution</B> <pre> {{{2sin^2(x) - 7 sin x + 3}}} = {{{0}}}. Group the terms to get factoring: {{{2sin^2(x) - 6sin(x) - sin(x) + 3}}} = {{{0}}}. {{{2sin(x)*(sin(x)-3) - (sin(x)-3)}}} = {{{0}}}, (sin(x)-3)*(2sin(x)-1) = 0. The last equation splits in two equations: 1. sin(x) = 3 (this equation has no solutions) or 2. sin(x) = 1/2 ---> x = {{{pi/6 + 2k*pi}}} or x = {{{(5/6)pi + 2k*pi}}}, where k is any integer k = 0, +/-1, +/-2, . . . <B>Answer</B>. The solutions are x = {{{pi/6 + 2k*pi}}} or x = {{{(5/6)pi + 2k*pi}}}, where k is any integer. </pre> <H3>Problem 3</H3>Solve an equation {{{2cos^2(theta)}}} = {{{sin(theta)+1}}}. <B>Solution</B> <pre> {{{2cos^2(theta)}}} = {{{sin(theta)+1}}}. Replace {{{cos^2(theta)}}} by {{{1 - sin^2(theta)}}} to get the equation for sine only. You will get {{{2*(1-sin^2(theta))}}} = {{{sin(theta)+1}}}, or {{{2 - 2sin^2(theta)}}} = {{{sin(theta)+1}}}, or {{{2*sin^2(theta) + sin(theta) - 1}}} = {{{0}}}, or (after factoring left side) {{{(2*sin(theta) - 1)*(sin(theta) + 1)}}} = {{{0}}}. The last equation splits in two equations: 1. {{{2*sin(theta) - 1}}} = {{{0}}} ---> {{{sin(theta)}}} = {{{1/2}}} ---> {{{theta}}} = {{{pi/6 + 2k*pi}}} or {{{theta}}} = {{{5pi/6 + 2k*pi}}}, k = 0, +/-1, +/-2, . . . 2. {{{sin(theta) + 1}}} = {{{0}}} ---> {{{sin(theta)}}} = {{{-1}}} ---> {{{theta}}} = {{{3pi/4 + 2k*pi}}} = {{{(2k+1)*pi}}}, k = 0, +/-1, +/-2, . . . <B>Answer</B>. The solution of the original equation is the union of the solutions to n.1 and n.2. </pre> <H3>Problem 4</H3>Solve an equation {{{3cos(beta) + 3}}} = {{{2sin^2(beta)}}}. <B>Solution</B> <pre> {{{3*cos(beta) + 3}}} = {{{2sin^2(beta)}}}. Replace {{{sin^2(beta)}}} by {{{1 - cos^2(beta)}}} to get the equation for cosine only. You will get {{{3*cos(beta) + 3}}} = {{{2*(1 - cos^2(beta))}}}, or {{{3*cos(beta) + 3}}} = {{{2 - 2*cos^2(beta)}}}, or {{{2*cos^2(beta) + 3*cos(beta) + 1}}} = {{{0}}}, or (after factoring left side) {{{(2*cos(beta) + 1)*(cos(beta) + 1)}}} = {{{0}}}. It splits in two equations: 1. {{{2*cos(beta) + 1}}} = {{{0}}} ---> {{{cos(beta)}}} = {{{-1/2}}} ---> {{{beta}}} = {{{2pi/3 + 2k*pi}}} or beta = {{{4pi/3 + 2k*pi}}}, k = 0, +/-1, +/-2, . . . 2. {{{cos(beta) + 1}}} = {{{0}}} ---> {{{cos(beta)}}} = {{{-1}}} ---> {{{beta}}} = {{{pi + 2k*pi}}} = {{{(2k+1)*pi}}}, k = 0, +/-1, +/-2, . . . <B>Answer</B>. The solution of the original equation is the union of the solutions to n.1 and n.2. </pre> <H3>Problem 5</H3>Solve an equation {{{cos(2x) + 6sin^2(x)}}} = {{{4}}}. <B>Solution</B> <pre> {{{cos(2x) + 6sin^2(x)}}} = {{{4}}}. The way to solve it is to transform the original equation into the equation for sine only. To do it, use the formula for cosine of doubled argument cos(2x) = {{{cos^2(x) - sin^2(x)}}} and replace {{{cos^2(x)}}} by {{{1-sin^2(x)}}}. You will get cos(2x) = {{{1 - 2*sin^2(x)}}}. Now substitute it into original equation. You will get {{{1 - 2*sin^2(x) + 6*sin^2(x)}}} = {{{4}}}. Collect like terms. You will get {{{4*sin^2(x)}}} = {{{3}}}, or {{{sin^2(x)}}} = {{{3/4}}}, or (after taking the square root from both sides) sin(x) = +/- {{{sqrt(3)/2}}}. The solutions are x = {{{pi/3}}}, {{{2pi/3}}}, {{{4pi/3}}}, {{{5pi/3}}}. In the interval [{{{0}}},{{{2pi}}}) these are all solutions. 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