{"id":2011,"date":"2025-07-31T23:43:52","date_gmt":"2025-07-31T23:43:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2011"},"modified":"2025-10-16T17:57:30","modified_gmt":"2025-10-16T17:57:30","slug":"solving-trigonometric-equations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-learn-it-2\/","title":{"raw":"Solving Trigonometric Equations: Learn It 2","rendered":"Solving Trigonometric Equations: Learn It 2"},"content":{"raw":"<h2>Solve Trigonometric Equations Using a Calculator<\/h2>\r\nNot all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Use a calculator to solve the equation [latex]\\sin \\theta =0.8[\/latex], where [latex]\\theta [\/latex] is in radians.[reveal-answer q=\"522194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522194\"]\r\n<h3><\/h3>\r\nMake sure mode is set to radians. To find [latex]\\theta [\/latex], use the inverse sine function. On most calculators, you will need to push the 2<sup>ND<\/sup> button and then the SIN button to bring up the [latex]{\\sin }^{-1}[\/latex] function. What is shown on the screen is [latex]{\\sin}^{-1}[\/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\\sin }^{-1}\\left(0.8\\right)[\/latex], and press ENTER. Thus, to four decimals places,\r\n<p style=\"text-align: center;\">[latex]{\\sin }^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/p>\r\nThis is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]\/pi - 0.9273 \\approx 2.2143 [\/latex]\r\n\r\nThe general solution is\r\n<p style=\"text-align: center;\">[latex]\\theta \\approx 0.9273\\pm 2\\pi k \\text{ and } \\theta \\approx 2.2143 \\pm 2\\pi k[\/latex]<\/p>\r\nThe angle measurement in degrees is\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\theta &amp;\\approx {53.1}^{\\circ } \\\\ \\theta &amp;\\approx {180}^{\\circ }-{53.1}^{\\circ } \\\\ &amp;\\approx {126.9}^{\\circ } \\end{align}[\/latex]<\/p>\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nNote that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\\pi -\\theta [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use a calculator to solve the equation [latex]\\sec \\theta =-4[\/latex], giving your answer in radians.[reveal-answer q=\"209133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"209133\"]We can begin with some algebra.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sec \\theta =-4\\\\ \\frac{1}{\\cos \\theta }=-4\\\\ \\cos \\theta =-\\frac{1}{4}\\end{gathered}[\/latex]<\/p>\r\nCheck that the MODE is in radians. Now use the inverse cosine function.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)\\approx 1.8235 \\\\ \\theta \\approx 1.8235+2\\pi k \\end{gathered}[\/latex]<\/p>\r\nSince [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14[\/latex], 1.8235 is between these two numbers, thus [latex]\\theta \\approx \\text{1}\\text{.8235}[\/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164111\/CNX_Precalc_Figure_07_05_0052.jpg\" alt=\"Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597\" width=\"487\" height=\"380\" \/> <b>Figure 2.<\/b>[\/caption]\r\n\r\nSo, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597}\\text{.}[\/latex]\r\n\r\nThe solutions are [latex]\\theta \\approx 1.8235\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.4597\\pm 2\\pi k[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<div class=\"bcc-box bcc-success\"><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nSolve [latex]\\cos \\theta =-0.2[\/latex].\r\n\r\n[reveal-answer q=\"145806\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145806\"]\r\n\r\n[latex]\\theta \\approx 1.7722\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.5110\\pm 2\\pi k[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]149873[\/ohm_question]<\/section><\/div>","rendered":"<h2>Solve Trigonometric Equations Using a Calculator<\/h2>\n<p>Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Use a calculator to solve the equation [latex]\\sin \\theta =0.8[\/latex], where [latex]\\theta[\/latex] is in radians.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q522194\">Show Solution<\/button><\/p>\n<div id=\"q522194\" class=\"hidden-answer\" style=\"display: none\">\n<h3><\/h3>\n<p>Make sure mode is set to radians. To find [latex]\\theta[\/latex], use the inverse sine function. On most calculators, you will need to push the 2<sup>ND<\/sup> button and then the SIN button to bring up the [latex]{\\sin }^{-1}[\/latex] function. What is shown on the screen is [latex]{\\sin}^{-1}[\/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\\sin }^{-1}\\left(0.8\\right)[\/latex], and press ENTER. Thus, to four decimals places,<\/p>\n<p style=\"text-align: center;\">[latex]{\\sin }^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/p>\n<p>This is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]\/pi - 0.9273 \\approx 2.2143[\/latex]<\/p>\n<p>The general solution is<\/p>\n<p style=\"text-align: center;\">[latex]\\theta \\approx 0.9273\\pm 2\\pi k \\text{ and } \\theta \\approx 2.2143 \\pm 2\\pi k[\/latex]<\/p>\n<p>The angle measurement in degrees is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\theta &\\approx {53.1}^{\\circ } \\\\ \\theta &\\approx {180}^{\\circ }-{53.1}^{\\circ } \\\\ &\\approx {126.9}^{\\circ } \\end{align}[\/latex]<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\\pi -\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use a calculator to solve the equation [latex]\\sec \\theta =-4[\/latex], giving your answer in radians.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q209133\">Show Solution<\/button><\/p>\n<div id=\"q209133\" class=\"hidden-answer\" style=\"display: none\">We can begin with some algebra.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sec \\theta =-4\\\\ \\frac{1}{\\cos \\theta }=-4\\\\ \\cos \\theta =-\\frac{1}{4}\\end{gathered}[\/latex]<\/p>\n<p>Check that the MODE is in radians. Now use the inverse cosine function.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)\\approx 1.8235 \\\\ \\theta \\approx 1.8235+2\\pi k \\end{gathered}[\/latex]<\/p>\n<p>Since [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14[\/latex], 1.8235 is between these two numbers, thus [latex]\\theta \\approx \\text{1}\\text{.8235}[\/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164111\/CNX_Precalc_Figure_07_05_0052.jpg\" alt=\"Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597\" width=\"487\" height=\"380\" \/><figcaption class=\"wp-caption-text\"><b>Figure 2.<\/b><\/figcaption><\/figure>\n<p>So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597}\\text{.}[\/latex]<\/p>\n<p>The solutions are [latex]\\theta \\approx 1.8235\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.4597\\pm 2\\pi k[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-success\">\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>Solve [latex]\\cos \\theta =-0.2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q145806\">Show Solution<\/button><\/p>\n<div id=\"q145806\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\theta \\approx 1.7722\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.5110\\pm 2\\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm149873\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149873&theme=lumen&iframe_resize_id=ohm149873&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n","protected":false},"author":13,"menu_order":27,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":201,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2011"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2011\/revisions"}],"predecessor-version":[{"id":4690,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2011\/revisions\/4690"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/201"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2011\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2011"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2011"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2011"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2011"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}